[POJ 3040] Allowance （贪心）

POJ - 3040

FJ手中有若干面值的货币，小面值的货币能被大面值的整除
每周他要给奶牛发不少于 C的工资，问最多能发几周

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首先大于 C的面值都先发掉
然后优先发剩下的大面额的货币，当超过时
将最后一个大面额的用尽可能小的货币代替

#pragma comment(linker, "/STACK:102400000,102400000")#include <cstdio>#include <iostream>#include <cstdlib>#include <cstring>#include <algorithm>#include <cmath>#include <map>#include <set>#include <queue>using namespace std;typedef pair<int,int> Pii;typedef long long LL;typedef unsigned long long ULL;typedef double DBL;typedef long double LDBL;#define MST(a,b) memset(a,b,sizeof(a))#define CLR(a) MST(a,0)#define Sqr(a) (a*a)const int INF=0x3f3f3f3f;struct data{    int v,b;    bool operator < (const data &y) const {return v < y.v;}} inpt[30];int N,C;int have[30];inline void pause(){fflush(stdin); getchar();}int main(){    #ifdef LOCAL    freopen("in.txt", "r", stdin);//  freopen("out.txt", "w", stdout);    #endif    while(~scanf("%d%d", &N, &C))    {        for(int i=0; i<N; i++) scanf("%d%d", &inpt[i].v, &inpt[i].b);        sort(inpt, inpt+N);        LL ans=0;        while(N && inpt[N-1].v >= C) ans += inpt[--N].b;        while(1)        {            for(int i=0; i<N; i++) have[i]=inpt[i].b;            int left=C;            for(int i=N-1; i>=0; i--)            {                int cnt = left/inpt[i].v;                if(have[i] < cnt) continue;                have[i] -= cnt;                left -= cnt*inpt[i].v;            }            for(int i=0; i<N; i++)            {                int cnt = left/inpt[i].v;                if(left%inpt[i].v) cnt++;                if(have[i] < cnt) cnt = have[i];                have[i] -= cnt;                left -= cnt*inpt[i].v;                if(left <= 0) break;            }            if(left > 0) break;            int tims=INF;            for(int i=0; i<N; i++) if(have[i] < inpt[i].b) tims = min(tims, inpt[i].b/(inpt[i].b - have[i]));            ans+=tims;            for(int i=0; i<N; i++) if(have[i] < inpt[i].b) inpt[i].b -= (inpt[i].b - have[i])*tims;        }        printf("%lld\n", ans);    }    return 0;}