Charlie's Change POJ - 1787

題目傳送門

題意：給你四種硬幣，硬幣的面值分別爲1，5，10，25，每一種硬幣分別有C1，C2，C3，C4，問你能不能通過這些硬幣構成p元，
（儘可能的多用硬幣）如果能的話輸出方案。

思路：完全揹包記錄一下路徑就好了。

#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

#define MAXN 10100
#define MAXE 10
#define INF 100000000
#define MOD 10001
#define LL long long
#define pi 3.14159

using namespace std;

int num[MAXN];
int v[MAXE];
int dp[MAXN];
int used[MAXN];
int path[MAXN][MAXE];

int main() {
    std::ios::sync_with_stdio(false);
    LL p;
    while (cin >> p >> num[1] >> num[2] >> num[3] >> num[4]) {
        if (p == 0 && num[1] == 0 && num[2] == 0 && num[3] == 0 && num[4] == 0)
            break;
        memset(path, 0, sizeof(path));
        v[1] = 1;
        v[2] = 5;
        v[3] = 10;
        v[4] = 25;
        dp[0] = 0;
        for (int i = 1; i < MAXN; ++i) {
            dp[i] = -INF;
        }
        for (int i = 1; i <= 4; ++i) {
            memset(used, 0, sizeof(used));
            for (int j = v[i]; j <= p; ++j) {
                if (dp[j] < dp[j - v[i]] + 1 && used[j - v[i]] < num[i]) {
                    dp[j] = dp[j - v[i]] + 1;
                    used[j] = used[j - v[i]] + 1;
                    for (int k = 1; k <= 4; ++k) {
                        path[j][k] = path[j - v[i]][k];
                    }
                    path[j][i]++;
                }
            }
        }
        if (dp[p] < 0) {
            cout << "Charlie cannot buy coffee.\n";
        } else {
            cout << "Throw in " << path[p][1] << " cents, " << path[p][2] << " nickels, " << path[p][3] << " dimes, and " << path[p][4] << " quarters.\n";
        }
    }
    return 0;
}

/*

 */