#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
const int INF=0x3f3f3f3f;#define met(a,b) (memset(a,b,sizeof(a)))
#define N 11000
#define INF 0x3f3f3f3fint pre[N], v[5]={0,1,5,10,25}, dp[N];
int used[N];/**dp[i]   代表组成i元时需要的最多的硬币
used[i] 代表第i种硬币用了几次（感觉这点很好，能将多重背包转化为完全背包）
pre[j]  记录的是j从哪个状态转化过来的*/int main()
{int p, num[5]={0};while(scanf("%d%d%d%d%d", &p, &num[1], &num[2], &num[3], &num[4]), p+num[1]+num[2]+num[3]+num[4]){int i, j, ans[110]={0};met(dp, -1);met(pre, -1);dp[0] = 0;for(i=1; i<=4; i++){memset(used, 0, sizeof(used));for(j=v[i]; j<=p; j++){if(dp[j-v[i]]+1>dp[j] && dp[j-v[i]]>=0 && used[j-v[i]]<num[i]){dp[j] = dp[j-v[i]]+1;used[j] = used[j-v[i]]+1;pre[j] = j-v[i];}}}if(dp[p]<0)printf("Charlie cannot buy coffee.\n");else{met(ans, 0);i = p;while(1){if(pre[i]==-1) break;ans[i-pre[i]]++;i = pre[i];}printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n", ans[v[1]], ans[v[2]], ans[v[3]], ans[v[4]]);}}return 0;
}

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>using namespace std;
#define met(a,b) (memset(a,b,sizeof(a)))
const int N=10001;
int dp[N], v[5]={0,1,5,10,25};
int num[N][5];//dp[j]为咖啡的价格为j时，所能花费的最多钱币数//num[j][i]，表示咖啡的价格为j时，花费的第i种货币的个数
int main()
{int a[5], p;while(scanf("%d%d%d%d%d", &p, &a[1], &a[2], &a[3], &a[4]), p+a[1]+a[2]+a[3]+a[4]){int i, j, k;met(dp, -1);met(num, 0);dp[0] = 0;for(i=1; i<=4; i++){for(j=v[i]; j<=p; j++){if(dp[j-v[i]]>=0 && dp[j-v[i]]+1>dp[j] && num[j-v[i]][i]<a[i]){dp[j] = dp[j-v[i]] + 1;for(k=1; k<=4; k++){if(k==i)num[j][k] = num[j-v[i]][k]+1;elsenum[j][k] = num[j-v[i]][k];}}}}if(dp[p]==-1)printf("Charlie cannot buy coffee.\n");elseprintf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n", num[p][1], num[p][2], num[p][3], num[p][4]);}return 0;
}