POJ

题意：你有1，5，10，25四种硬币数量有限，你想买价格为p的咖啡，问你能不能购买，要求花的硬币尽量多，并且输出每种硬币花了多少；
思路：看起来像多重背包加记录路径但用完全背包比较好写

#include<iostream>#include<algorithm>#include<string>#include<cstring>#include<map>#include<queue>#include<cmath>#include<stack>#include<vector>#include<cstdio>#define MAXN 33000#define INF 0x3f3f3f3f#define lmid l,m,rt<<1#define rmid m+1,r,rt<<1|1#define ls rt<<1#define rs rt<<1|1#define Mod 1000000007#define i64 __int64#define LIMIT_ULL 100000000000000000using namespace std;int dp[10005];int a[10005][4];int s[4];int t[4]={1,5,10,25};int main(){     int n;     while(scanf("%d",&n)==1)     {          memset(dp,-1,sizeof(dp));          memset(a,0,sizeof(a));          for(int i=0;i<4;i++)          {               scanf("%d",&s[i]);          }          if(n==0&&s[0]==0&&s[1]==0&&s[2]==0&&s[3]==0)               break;          dp[0]=0;          for(int i=0;i<4;i++)               for(int j=t[i];j<=n;j++)          {               if(dp[j-t[i]]!=-1)               {                    if(dp[j]<dp[j-t[i]]+1&&a[j-t[i]][i]+1<=s[i])                    {                         dp[j]=dp[j-t[i]]+1;                         for(int k=0;k<4;k++)                              a[j][k]=a[j-t[i]][k];                         a[j][i]++;                    }               }          }          if(dp[n]==-1)          {               cout<<"Charlie cannot buy coffee."<<endl;               continue;          }          printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n",a[n][0],a[n][1],a[n][2],a[n][3]);     }     return 0;}