1

# ICPC 2019 Preli: G – Pairs Forming GCD

0

Solution Idea:

• We have to find maximum G such that there exist at least P pairs of integers (X, Y), where 1 ≤ X ≤ Y ≤ N and GCD(X, Y) = G.
• At first, let’s try to solve a subproblem where G=1. Can we calculate the number of pairs (X, Y) where 1 ≤ X ≤ Y ≤ N and GCD(X, Y) = 1?
• As we know Euler totient or Phi function gives us a number of X which are coprime with Y. So Phi(Y) = number of X such that GCD(X,Y) = 1.
• So Phi(1)+Phi(2)+Phi(3)+…….+Phi(N) = number of pair (X,Y) where 1 ≤ X ≤ Y ≤ N and GCD(X, Y) = 1.
• Now if GCD(X,Y) = 1 then GCD(GX,GY) = G.
• So we can pre-calculate Euler totient for 1 to N and then calculate the cumulative sum of this phi[] array and store it to phiSum[] array.
• After this, we just have to binary search for the maximum answer on the new phiSum[] array.

```#include <bits/stdc++.h>

using namespace std;

#define mx 10000007
int phi[mx];
long long phiSum[mx];

void euler_totient() {
for(int i=0; i<mx; i++) phi[i]=i;
for(int i=2; i<mx; i++) {
if(phi[i]==i) {
for(int j=i; j<mx; j+=i)
phi[j]-=(phi[j]/i);
}
}
}

int solve(int n, long long p) {
int lo=1,hi=n;
int ret=-1;
while(lo<=hi) {
int mid=(lo+hi)/2;
if(phiSum[mid]>=p) {
ret=mid;
hi=mid-1;
} else {
lo=mid+1;
}
}
return ret;
}

int main() {

euler_totient();
for(int i=1; i<mx; i++) phiSum[i]+=phiSum[i-1]+phi[i]; //cumulative sum

int t;
scanf("%d",&t);
for(int z=1; z<=t; z++) {
long long n,p;
scanf("%lld %lld",&n,&p);
int ans=solve(n,p);
printf("Case %d: ",z);
if(ans!=-1)
ans=n/ans;
printf("%d\n",ans);
}
return 0;
}```

# Light OJ: 1144 – Ray Gun

0
Solution Idea:

• There are many observations to make in order to get to a working solution.
• For every lattice point (i, j), the ray that intersects it is unique and it’s identified by the pair , where g is the gcd of i and j.
• The problem is now reduced to counting the number of irreducible fractions such that a ≤ N and b ≤ M. This is the same as counting for every i between 1 and N, the amount of numbers in the range [1, M] that are coprime with i.
• Consider a certain number x with prime factors p1, p2. How do we know how many numbers in range [1, M] are coprime with it? That’s equal to M minus the amount of multiples of p1 minus the amount of multiples of p2 plus the amount of multiples of p1 * p2. This is inclusion-exclusion, and in general, if the amount of elements is even, we add, otherwise, we subtract.
• So now we have a working (but slow) solution: Iterate over every i in the range [1, N] and for every i, factorize it, try out all combinations of primes and then, for every combination that results in a number k, add if the amount of primes is even or subtract if the amount of primes is odd.
• The previous approach is very slow for two reasons: You’ll be factorizing each number every time and you’ll be doing a lot of repeated work. Every combination of primes you try out at each step will result in a certain number k. A crucial observation is that the higher exponent of that number k will be 1, because we’re trying combinations of different primes. Another crucial observation is that this number k will be seen times in total. Finally, each time we see it, it will contribute by to the final answer (or if the amount of primes is odd).
• Knowing all this, we can precalculate a lot of stuff and then solve each test case in O(N). We should precalculate the amount of prime factors of every number in the range [1, 106] (this can be done with a simple sieve), and we should cross out numbers that have some prime with an exponent higher than 1 (in other words, multiples of some square). Once we have precalculated all that, we simply iterate from 1 to N and for every number x that we didn’t cross out, we add (or subtract) to our answer.
• Final observations: We should add 2 to our answer (the two borders). If N = 0, the answer is 1, except M = 0 too, in which case the answer is 0.

(This solution idea is from this link )

# Light OJ: 1170 – Counting Perfect BST

0

```
#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              100000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;

/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

ll numbers;
int cnt=0;
void generate_num()
{
for(ll i=2;i<=100000;i++)
{
ll num=i*i;
while(num<=10000000000)
{
numbers[cnt++]=num;
num*=i;
}
}

sort(numbers,numbers+cnt);
cnt=unique(numbers,numbers+cnt)-numbers;
numbers[cnt++]=1000000000000000;

}

ll fact;

void gen_fact()
{
fact=1;
for(int i=1;i<=1000005;i++)
{
fact[i]=(fact[i-1]*i)%MOD;
}
}

ll mod_inv(ll n, ll pow)
{
if(pow==0) return 1;
if(pow%2==0)
{
ll ret=mod_inv(n,pow/2)%MOD;
return (ret*ret)%MOD;
}
return ((n%MOD)*(mod_inv(n,pow-1)%MOD))%MOD;
}

int main()
{

///freopen("in.txt","r",stdin);
///freopen("out.txt","w",stdout);

generate_num();
gen_fact();

int t;
sf(t);
TEST_CASE(t)
{
ll a,b;
sffl(a,b);

ll r=upper_bound(numbers,numbers+cnt,b)-numbers;
ll l=lower_bound(numbers,numbers+cnt,a)-numbers;
ll n=(r-l);
PRINT_CASE;
if(n==0)
pf("0\n");
else
{
ll ans=(fact[n+1]*fact[n])%MOD;
ans=mod_inv(ans,MOD-2);
ans=(fact[2*n]*ans)%MOD;
pf("%lld\n",ans);
}

}

return 0;
}

```

# UVa 10139 – Factovisors

0

Solution Idea:

Calculate the prime factor of M. Let a prime is P and it’s power is x. so P^x is a prime factor of M. then check that in N! prime factor of P is grater or equal to x.

```

#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;

/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

#define mx 100000

bitset<mx/2> vis;

vector<int>prime;

void sive()
{
int x=mx/2,y=sqrt(mx)/2;

for(int i=1; i<y; i++)
{
if(vis[i]==0)
{
for(int j=i*(i+1)*2; j<x; j+=(2*i)+1)
vis[j]=1;
}
}

prime.pb(2);
for(int i=3; i<mx; i+=2)
if(vis[i/2]==0)
prime.pb(i);

}

int main()
{
CIN;
//    freopen("in.txt","r",stdin);
///freopen("out.txt","w",stdout);

sive();

ll n,m;
while(cin>>n>>m)
{
ll a,b;
a=n,b=m;
ll root=sqrt(m)+1;
bool test=0;
//        if(m==0) test=1;
for(int i=0;prime[i]<root;i++)
{
if(m%prime[i]==0)
{
int cnt=0;
while(m%prime[i]==0)
cnt++,m/=prime[i];
ll temp=n;
ll sum=0;
while(temp)
{
sum+=temp/prime[i];
temp/=prime[i];
}
if(sum<cnt)
{
test=1;
break;
}
root=sqrt(m)+1;
}
}
if(m>1)
{
ll temp=n;
ll sum=0;
while(temp)
{
sum+=temp/m;
temp/=m;
}
if(sum<1)
{
test=1;
}
}

if(test==0)
cout<<b<<" divides "<<a<<"!"<<endl;
else
cout<<b<<" does not divide "<<a<<"!"<<endl;
}

return 0;
}

```

# 10673 – Play with Floor and Ceil

0

Solution Idea:

Straight forward implementation.

Just calculate a and b of extended euclid algorithm and pass it to extended euclid function. don’t worry about sample input and output 😛

```

#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;

/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

ll extended_euclid(ll a, ll b, ll &x, ll &y)
{
if(b==0)
{
x=1;
y=0;
return a;
}
ll x1,y1;
ll ret=extended_euclid(b,a%b,x1,y1);
x=y1;
y=x1-y1*(a/b);
return ret;
}

int main()
{

///freopen("in.txt","r",stdin);
///freopen("out.txt","w",stdout);

int t;
cin>>t;
TEST_CASE(t)
{
ll x,k;
cin>>x>>k;
ll a=x/k;
ll b=(x/k)+ ((x%k)?1:0);

ll xx,yy;

ll gcdd=extended_euclid(a,b,xx,yy);
x/=gcdd;
xx*=x;
yy*=x;
cout<<xx<<" "<<yy<<endl;

}

return 0;
}

```

# UVa 10090 – Marbles

0

Solution Idea: first check if it is possible or not to get a valid result. If possible then get one result with extended euclid algo and then check the result taking minimum type 1 box and then check the result taking minimum type 2 box. print the minimum value.

```
#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0);cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;

/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

ll extended_euclid(ll a, ll b, ll &x, ll &y)
{
if(b==0)
{
x=1,y=0;
return a;
}
ll x1,y1;
ll ret=extended_euclid(b,a%b,x1,y1);
x=y1;
y=x1-y1*(a/b);
return ret;
}

int main()
{
CIN;
//    freopen("in.txt","r",stdin);
///freopen("out.txt","w",stdout);

ll n;
while(cin>>n && n)
{
ll c1,n1,c2,n2;
cin>>c1>>n1>>c2>>n2;

bool test=0;

if(n % __gcd(n1,n2))
{
cout<<"failed"<<endl;
}
else
{
ll x,y;
ll gcdd=extended_euclid(n1,n2,x,y);
x*=(n/gcdd);
y*=(n/gcdd);

n1/=gcdd;
n2/=gcdd;

ll aa=ceil(-(double)x/n2);
ll bb=floor((double)y/n1);

if(aa<=bb)
{
if(c1 * (x + n2 * aa) + c2 * (y- n1*aa) < c1*(x+(n2*bb)) + c2*(y-n1*bb))
{
x+=n2*aa;
y-=n1*aa;
}
else
{
x+=n2*bb;
y-=n1*bb;
}
}
else
test=1;

if(test)
cout<<"failed"<<endl;
else
cout<<x<<" "<<y<<endl;

}

}

return 0;
}

```

# UVa 10104 – Euclid Problem

0

Solution Idea: Straight forward implementation of extended euclid algo.

```
#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;

/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

ll extended_euclid(ll a, ll b, ll &x, ll &y)
{
if(b==0)
{
x=1;y=0;
return a;
}
ll x1,y1;
ll temp=extended_euclid(b,a%b,x1,y1);
x=y1;
y=x1-y1*(a/b);
return temp;
}

int main()
{

///freopen("in.txt","r",stdin);
///freopen("out.txt","w",stdout);

ll a,b;
while(sffl(a,b)==2)
{
ll x,y,z;
z=extended_euclid(a,b,x,y);
pf("%lld %lld %lld\n",x,y,z);
}

return 0;
}

```

# Light OJ: 1340 – Story of Tomisu Ghost

2

Details about Prime Factorization of Factorial : http://goo.gl/qbNxfH

```
/*
If opportunity doesn't knock, build a door.

+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
|S|.|S|.|R|.|A|.|S|.|A|.|M|.|K|
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+

Success is how high you bounce when you hit bottom.
*/

#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              10000019
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;

/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

#define maxx 100005

bitset<maxx/2>vis;
vector<int>prime;

void sieve()
{
int x=maxx/2, y=sqrt(maxx)/2;
for(int i=1;i<=y;i++)
{
if(vis[i]==0)
{
for(int j=i*(i+1)*2;j<=x;j+=(2*i)+1)
vis[j]=1;
}
}
prime.pb(2);
for(int i=3;i<maxx;i+=2)
if(vis[i/2]==0)
prime.pb(i);
}

ll bigmod(ll n, ll pow, ll mod)
{
if(pow==0) return 1;
if(pow%2==0)
{
ll ret=bigmod(n,pow/2,mod)%mod;
return (ret*ret)%mod;
}
return ((n%mod)*(bigmod(n,pow-1,mod)%mod))%mod;
}

ll factorial_factor(int n, int t)
{
ll ans=1;

for(int i=0;i<SZ(prime) && prime[i]<=n;i++)
{
int nn=n;
ll sum=0;
while(nn/prime[i])
{
sum+=nn/prime[i];
nn/=prime[i];
}
if(sum>=t)
{
sum/=t;
ans*=bigmod(prime[i],sum,MOD);
ans%=MOD;
}
}
return ans;
}

int main()
{

///freopen("in.txt","r",stdin);
///freopen("out.txt","w",stdout);

sieve();
int t;
sf(t);
TEST_CASE(t)
{
int a,b;
sff(a,b);
ll xx=factorial_factor(a,b);
PRINT_CASE;
if(xx==1)
pf("-1\n");
else
pf("%lld\n",xx);
}

return 0;
}

```

# Light OJ: 1163 – Bank Robbery

0

```
/*

SOLUTION IDEA:
--------------

Let, the given number is X = A - B. Here, B = A/10.
So, A - A/10 = X
A - (A-A%10)/10 = X
10A - A + (A%10) = 10X
9A = 10X - K , let K = A%10
A = (10X - K)/9
A = X + (X - K)/9
For K equals to 0 to 9, if (X - K)%9 = 0, then A would be a solution.
If we get a solution for K = 0, then we would also get a solution for
K = 9 in this case. That means, if X%9 = 0, then there exists two solution.

*/

/*
If opportunity doesn't knock, build a door.

+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+
|S|.|S|.|R|.|A|.|S|.|A|.|M|.|K|
+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+

Success is how high you bounce when you hit bottom.
*/

#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;

/*----------------------Graph Moves----------------*/
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

int main()
{

///freopen("in.txt","r",stdin);
///freopen("out.txt","w",stdout);

int t;
sf(t);
TEST_CASE(t)
{
ll a;
sfl(a);
bool test=0;
PRINT_CASE;
for(int i=9;i>=0;i--)
{
if((a-i)%9==0)
{
if(test) printf(" ");
printf("%lld",a+((a-i)/9));
test=1;
}
}
printf("\n");
}

return 0;
}

```