Problem Link : http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=2512
Idea:
1. This is a classical coin change problem. here dp array represent the minimum coin need to make an amount. example – dp[100] means we need at least dp[100] coins to make 100 cents.
2. At first calculate the dp[] table as described. then start a linear search from money to forward for finding first amount of money which value is not infinity as we set before dp calculation and when we find a such value that value is the payable money and dp value of that money is the minimum number of coins.
/* MMP""MM""YMM db `7MN. `7MF'`7MMM. ,MMF' .g8""8q.`YMM' `MM' P' MM `7 ;MM: MMN. M MMMb dPMM .dP' `YM.VMA ,V MM ,V^MM. M YMb M M YM ,M MM dM' `MM VMA ,V MM ,M `MM M `MN. M M Mb M' MM MM MM VMMP MM AbmmmqMA M `MM.M M YM.P' MM MM. ,MP MM MM A' VML M YMM M `YM' MM `Mb. ,dP' MM .JMML..AMA. .AMMA..JML. YM .JML. `' .JMML. `"bmmd"' .JMML. */ #include <bits/stdc++.h> #define pii pair <int,int> #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 oo 1<<29 #define dd double #define ll long long #define EPS 10E-10 #define ff first #define ss second #define MAX 10000 #define CIN ios_base::sync_with_stdio(0); cin.tie(0) #define SZ(a) (int)a.size() #define si(a) scanf("%d",&a) #define sii(a,b) scanf("%d%d",&a,&b) #define siii(a,b,c) scanf("%d%d%d",&a,&b,&c) #define loop(i,n) for(int i=0;i<n;i++) #define REP(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 all(a) a.begin(),a.end() #define intlim 2147483648 #define inf 1000000 #define rtintlim 46340 #define llim 9223372036854775808 #define rtllim 3037000499 #define ull unsigned long long #define I int using namespace std; int dp[20100]; int coin[110]; int main() { ///freopen("in.txt","r",stdin); ///freopen("out.txt","w",stdout); int t; si(t); TEST_CASE(t) { int money; si(money); int n; si(n); loop(i,n) { si(coin[i]); } //sort(coin,coin+n,test); ms(dp,inf); dp[0]=0; for(int i=0; i<n; i++) { for(int j=money; j>-1; j--) { dp[j+coin[i]]=min(dp[j]+1,dp[j+coin[i]]); } } for(int i=money; i<=30100; i++) { if(dp[i]<10000) { pf("%d %d\n",i,dp[i]); break; } } } return 0; }
Another Solution using same idea. this one is from revise 😛 .
/* +-+ +-+ +-+ |R| |.| |S| +-+ +-+ +-+ */ #include <bits/stdc++.h> #define pii pair <int,int> #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 100007 #define MAX 10000 #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 loop(i,n) for(int i=0;i<n;i++) #define REP(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 all(a) a.begin(),a.end() #define intlim 2147483648 #define ull unsigned long long 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 /*------------------------------------------------*/ /*-----------------------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));} /*------------------------------------------------*/ int dp[1000007]; int coin[110]; int main() { ///freopen("in.txt","r",stdin); ///freopen("out.txt","w",stdout); int t; sf(t); TEST_CASE(t) { ms(dp,1000000); int money,n; sf(money); sf(n); loop(i,n) sf(coin[i]); dp[0]=0; for(int i=0; i<n; i++) { for(int j=money; j>=0; j--) { dp[j+coin[i]]=min(dp[j]+1,dp[j+coin[i]]); } } for(int i=money;; i++) { if(dp[i]<1000000 && dp[i]) { pf("%d %d\n",i,dp[i]); break; } } } return 0; }