poj1787

//由于此题是要求恰好装满，故初始化时为-OO#include<iostream>#include<cstdio>#include<cstring>using namespace std;#define maxn 10005#define INF -0xffffffint v;int f[maxn];int lm[4],w[4];int sum[maxn][4];void zero(int c,int num,int id) //第id种硬币{    for(int i=v;i>=c;i--)    {        if(f[i]<f[i-c]+num&&f[i-c]!=INF)        {            f[i]=f[i-c]+num;            for(int j=0;j<4;j++)                sum[i][j]=sum[i-c][j];            sum[i][id]+=num;        }    }}void com(int c,int id) //第id种硬币{    for(int i=c;i<=v;i++)    {        if(f[i]<f[i-c]+1&&f[i-c]!=INF)        {            f[i]=f[i-c]+1;            for(int j=0;j<4;j++)                sum[i][j]=sum[i-c][j];            sum[i][id]+=1;        }    }}int main(){    int w[4]={1,5,10,25};    while(scanf("%d%d%d%d%d",&v,&lm[0],&lm[1],&lm[2],&lm[3]))    {        if(v+lm[0]+lm[1]+lm[2]+lm[3]==0) break;        for(int i=0;i<maxn;i++)            f[i]=INF;        f[0]=0;        memset(sum,0,sizeof(sum));        for(int i=0;i<4;i++)        {            if(w[i]*lm[i]>=v)                com(w[i],i);            else{                for(int k=1;k<lm[i];k=k*2) //二进制分解                {                    zero(w[i]*k,k,i);                    lm[i]-=k;                }                zero(w[i]*lm[i],lm[i],i);            }        }        if(f[v]==INF)            printf("Charlie cannot buy coffee.\n");        else            printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",sum[v][0],sum[v][1],sum[v][2],sum[v][3]);    }    return 0;}//f[i][j]表示前i种硬币放入体积为j的背包所花费的硬币的最大枚数//sum[i][4]表示组成体积为i的背包所需要的四种硬币的数量