SPOJ: SUMSUM – Enjoy Sum with Operations

Problem Link : http://www.spoj.com/problems/SUMSUM/

Solution Idea:

    This one is a interesting problem. The main idea behind the problem is combinatorics. At first we need to count the number of 1 bit and number of 0 bit in the i_th position of a given number. Then we need to do some combinatorial calculation.

    Now think about an array of 4 number A=[1,0,1,0]. If we perform OR operation in the range 1 to 4 what will we get? We get 5 as answer. If we form pair from this number and perform OR operation between then then we will get the answer. Now as here is 4 element so we can perform pair in 4*(4-1)/2 = 6 ways. Now comes the tricky part. For OR operation how many pair will contribute nothing ?? Here is 2 zero and we can form 1 pair from this two element who will not contribute anything. So for i_th bit position the contribution for OR operation is number_of_active_pair*(2^i).

    For AND operation we need to consider the number of pair can be form by 1 bit and for XOR the number of pair will be number_of_one_bit*number_of_zero_bit.

    We can consider a number as an array of 30 element whose value is either 0 or 1 and perform above operation.



#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#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)             cerr<<#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 LINE_PRINT_CASE  printf("Case %d:\n",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;

//using namespace __gnu_pbds;
//typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;


/*----------------------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
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//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 100005

int tree[28][mx];

void update(int id, int idx, int val)
{
    for(;idx<mx && idx;idx+=idx&-idx)
        tree[id][idx]+=val;
}

ll query(int id, int idx)
{
    ll ret=0;
    for(;idx;idx-=idx&-idx)
        ret+=tree[id][idx];
    return ret;
}

int ara[mx];

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

    int n,q;
    sff(n,q);
    for(int i=1;i<=n;i++) sf(ara[i]);

    for(int i=1;i<=n;i++)
    {
        for(int j=0;j<28;j++)
        {
            if(check(ara[i],j))
                update(j,i,1);
        }
    }

    while(q--)
    {
        int a,b,c;
        sf(a);
        if(a==1)
        {
            sff(b,c);
            for(int i=0;i<28;i++)
            {
                if(check(ara,i))
                    update(i,c,-1);
            }
            ara=b;
            for(int i=0;i<28;i++)
            {
                if(check(ara,i))
                    update(i,c,1);
            }
        }
        else
        {
            char ss[10];
            scanf(" %s",&ss);
            string str=string(ss);
            sff(b,c);
            ll temp[30];
            for(int i=0;i<28;i++)
            {
                temp[i]=query(i,c);
            }
            for(int i=0;i<28;i++)
            {
                temp[i]-=query(i,b-1);
            }

            ll ans=0;

            for(int i=0;i<28;i++)
            {
                ll one=temp[i];
                ll zero=(c-b+1)-one;
                ll pairs=0;
                if(str=="OR")
                {
                    ll total=one+zero;
                    pairs=(total*(total-1))/2;
                    pairs-=(zero*(zero-1))/2;
                }
                else if(str=="AND")
                {
                    pairs=(one*(one-1))/2;
                }
                else if(str=="XOR")
                {
                    pairs=one*zero;
                }
                ans+=(1LL<<i)*pairs;
            }

            printf("%lld\n",ans);
        }
    }


    return 0;
}

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