POJ-1787 Charlie's Change (完全背包+输出方案 入门题)

Charlie's Change

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 4505		Accepted: 1420

Description

Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at coffee vending machines at motorests. Charlie hates change. That is basically the setup of your next task. 

Your program will be given numbers and types of coins Charlie has and the coffee price. The coffee vending machines accept coins of values 1, 5, 10, and 25 cents. The program should output which coins Charlie has to use paying the coffee so that he uses as many coins as possible. Because Charlie really does not want any change back he wants to pay the price exactly. 

Input

Each line of the input contains five integer numbers separated by a single space describing one situation to solve. The first integer on the line P, 1 <= P <= 10 000, is the coffee price in cents. Next four integers, C1, C2, C3, C4, 0 <= Ci <= 10 000, are the numbers of cents, nickels (5 cents), dimes (10 cents), and quarters (25 cents) in Charlie's valet. The last line of the input contains five zeros and no output should be generated for it.

Output

For each situation, your program should output one line containing the string "Throw in T1 cents, T2 nickels, T3 dimes, and T4 quarters.", where T1, T2, T3, T4 are the numbers of coins of appropriate values Charlie should use to pay the coffee while using as many coins as possible. In the case Charlie does not possess enough change to pay the price of the coffee exactly, your program should output "Charlie cannot buy coffee.".

Sample Input

12 5 3 1 2
16 0 0 0 1
0 0 0 0 0

Sample Output

Throw in 2 cents, 2 nickels, 0 dimes, and 0 quarters.
Charlie cannot buy coffee.

#include 
#include 
using namespace std;
int dp[10005], used[10005], pre[10005], coin[4] = {1, 5, 10, 25}, num[4], ans[26];
int main(){
	int p;
	while(scanf("%d %d %d %d %d", &p, &num[0], &num[1], &num[2], &num[3]) != EOF){
		if(p + num[0] + num[1] + num[2] + num[3] == 0) return 0;
		for(int i = 0; i <= p; ++i){
			dp[i] = -1e9;
		}
		memset(pre, 0, sizeof(pre));
		pre[0] = -1;
		dp[0] = 0;
		for(int i = 0; i < 4; ++i){
			memset(used, 0, sizeof(used));
			for(int j = coin[i]; j <= p; ++j){
				if(dp[j] < dp[j - coin[i]] + 1 && used[j - coin[i]] < num[i]){
					used[j] = used[j - coin[i]] + 1;
					dp[j] = dp[j - coin[i]] + 1;
					pre[j] = j - coin[i];
				}
			}
		}
		if(dp[p] <= 0){
			printf("Charlie cannot buy coffee.\n");
			continue;
		}
		memset(ans, 0, sizeof(ans));
		while(1){
			if(pre[p] == -1) break;
			ans[p - pre[p]]++;
			p = pre[p];
		}
		printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n", ans[1], ans[5], ans[10], ans[25]);
	}
}

/*
题意：10000元，4种面值硬币，给出每种硬币的个数，问最多可以用多少个硬币组合成给定的总价格，输出方案。

思路：就是完全背包需要输出方案，dp[j]表示总价格为j时最多可以有多少个硬币，但是需要维护一下硬币的个数，
实际上在01背包中，我们对所有状态都更新了一次答案，这里我们等于是对4种硬币各跑一次背包，维护一下使用
次数即可。used[j]表示总价格为j时用了多少个某个硬币。pre[j]记录方案，表示j出现更有的答案时是由哪个状态
转移来的。
最后遍历一下输出答案就好了。这里不能滚动数组来做，因为硬币不止1个，滚动只更新了一次答案。
*/