POJ1787——背包DP（特定状态+回溯）——Charlie's Change

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.
大意：给你一些钱，以及1.5.10.25零钱的个数，问你是否能拆开，并求出最多的分成的零钱的情况
第一次发现完全背包也能解决这种有限制的问题，一直以为只能在多重背包里面才能，原来用个if条件判断当前的num是否超过了t[i]就行
1.完全背包做法

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int v[5] = {0,1,5,10,25};
int dp[10010],ans[10010],num[10010],path[10010],t[5];
int p;
int main()
{
    while(~scanf("%d",&p)){
        for(int i = 1; i <= 4;  i++)
            scanf("%d",&t[i]);
        if((p+t[1]+t[2]+t[3]+t[4]) == 0) break;
        memset(dp,0,sizeof(dp));
        memset(ans,0,sizeof(ans));
        memset(path,0,sizeof(path));
        dp[0] = 1;
        for(int i = 1; i <= 4; i++){
            memset(num,0,sizeof(num));
            for(int j = v[i]; j <= p; j++){
                if(dp[j-v[i]] && dp[j-v[i]] + 1 > dp[j] && num[j-v[i]] < t[i]){
                    dp[j] = dp[j-v[i]] + 1;//使得后面每一个状态都是从前面一个得到，并且满足两个条件1：用去一个后的数目要比没用去的多2：用去的硬币数目不超过硬币本身
                    num[j] = num[j-v[i]] + 1;
                    path[j] = j - v[i];//难想到。。用来记录路径
                }
            }
        }
        int i = p;
        if(dp[p]>0){
            while(i!=0){
                ans[i-path[i]]++;//i-path[i] = i - ( i - v[i]) = v[i]
                i = path[i];//path[i]表示有i钱的时候用了一种硬币之后的钱的数目
            }
            printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.
",ans[1],ans[5],ans[10],ans[25]);
        }
        else printf("Charlie cannot buy coffee.
");
    }
    return 0;
}

　2.