描述

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.

輸入

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.

輸出

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.".

題目意思：有 4 種硬幣，給出咖啡的價格和四種硬幣的數量，求出能買到咖啡的最多硬幣數量，並輸出每一種硬幣對應的數量，如果買不到咖啡，就輸出不能買到。

用二進制優化的多重背包做，要記錄路徑。

dp[i] 表示 i 元錢時，硬幣最多是多少。

ans[i][j] 表示 i 元錢時，第 j 種硬幣的數量是多少。

在進行狀態轉移的時候，更新下 ans 數組，最後輸出 ans[p] 的四種硬幣數量就行了。

#include <iostream>
#include <cstring>
using namespace std;

const int MAX = 10500;

int dp[MAX];    //價值為 i 時，最多多少枚硬幣
int v[5] = {0, 1, 5, 10, 25};
int num[5];
int ans[MAX][5];    //價值為 i 時，四種硬幣的數量，此時硬幣數量和最多（等於dp[i]） 
int p;

void zeroOne(int i, int j);        //第 i 個物品，數量為 j 

int main(){
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    
    while(scanf("%d", &p)){
        int sum = 0;
        for(int i=1; i<=4; i++){
            scanf("%d", num+i);
            sum += num[i] * v[i];
        }
        
        if(sum == 0 && p == 0){
            break;
        }
        
        //錢不夠 
        if(sum < p){
            printf("Charlie cannot buy coffee.\n");
            continue;
        }
        
        //DP 多重背包 
        memset(dp, -1, sizeof(dp));
        memset(ans, 0, sizeof(ans)); 
        dp[0] = 0;
        for(int i=1; i<=4; i++){
            int j;
            for(j=1; j<=num[i]; j=(j<<1)){        //轉化成0-1背包進行二進制優化 
                zeroOne(i, j);
            }
            j = num[i];
            zeroOne(i, j);
        }
        
        if(dp[p] != -1){
            printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.\n", ans[p][1], ans[p][2], ans[p][3], ans[p][4]);
        }else{
            printf("Charlie cannot buy coffee.\n");
        }
        
    }
    
    return 0;
} 

void zeroOne(int i, int j){
    for(int k=p-j*v[i]; k>=0; k--){
        if(dp[k] != -1){                    //如果 k 狀態存在 
            if(dp[k] + j > dp[k+j*v[i]]){    //從 k 狀態出發，到 k + j*v[i] 狀態，如果所用硬幣比原來要多 
                int nk = k+j*v[i];
                dp[nk] = dp[k] + j;
                memcpy(ans[nk], ans[k], sizeof(ans[k]));    //更新記錄的硬幣數量 
                ans[nk][i] += j;        //相比 k 狀態多放了 j 個 i 
            }
        }
    }
    num[i] -= j;
}

臺州 OJ 2537 Charlie's Change 多重背包 二進制優化 路徑記錄