poj 1742

题意：给你n种面值的硬币，面值为a1…an，数量分别为c1…cn，求问，在这些硬币的组合下，能够多少种面值，该面值不超过m

然后代码是这样的

#include <iostream>
#include <cstdio>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>
#include <stdlib.h>
#include <stdio.h>

#define INF 0x3f3f3f3f
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define ull unsigned long long
#define uint unsigned int
#define l(x) (x<<1)
#define r(x) (x<<1|1)
#define ms(a,b) memset(a,b,sizeof(a))

using namespace std;

int a[1111], c[1111], dp[1111111],s[1111111];
int n, m,ans;

int main() {
	while (scanf("%d%d", &n, &m), n != 0) {
		for (int i = 0; i < n; i++) {
			scanf("%d", &a[i]);
		}
		for (int i = 0; i < n; i++) {
			scanf("%d", &c[i]);
		}

		ms(dp, -1);
		for (int i = 0; i < n; i++) {
			dp[0] = 1;
			for (int j = 0; j <= m; j++) {
				if (dp[j] > 0) {
					dp[j] = 1;
					s[j] = c[i];
				}
				else if (j >= a[i] && dp[j - a[i]] > 0 && s[j - a[i]] > 0) {
					dp[j] = 1;
					s[j] = s[j - a[i]] - 1;
				}
			}
		}

		ans = 0;
		for (int i = 1; i <= m; i++) {
			if (dp[i] > 0) ans++;
		}
		printf("%d\n", ans);
	}


	return 0;
}

然后可以发现，dp数组和s数组可以合并，稍微化简一下

#include <iostream>
#include <cstdio>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>
#include <stdlib.h>
#include <stdio.h>

#define INF 0x3f3f3f3f
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define ull unsigned long long
#define uint unsigned int
#define l(x) (x<<1)
#define r(x) (x<<1|1)
#define ms(a,b) memset(a,b,sizeof(a))

using namespace std;

int a[1111], c[1111], dp[1111111],s[1111111];
int n, m,ans;

int main() {
	while (scanf("%d%d", &n, &m), n != 0) {
		for (int i = 0; i < n; i++) scanf("%d", &a[i]);
		for (int i = 0; i < n; i++) scanf("%d", &c[i]);
		
		ms(dp, -1);
		for (int i = 0; i < n; i++) {
			dp[0] = 1;
			for (int j = 0; j <= m; j++) 
				if (dp[j] >= 0) dp[j] = c[i];
				else if (j >= a[i]) dp[j] = dp[j - a[i]] - 1;
		}

		ans = 0;
		for (int i = 1; i <= m; i++) if (dp[i] >= 0) ans++;
		printf("%d\n", ans);
	}


	return 0;
}