CDOJ 31 饭卡(card) 解题报告

题目链接:http://acm.uestc.edu.cn/#/problem/show/31

01背包一眼题,没什么好解释……

背包容量是m-5,留下一个最贵不要在01背包里算,最后买。

没错我需要<algorithm>

#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;

int n, m, p[1005];
int opt[1005];

int nextInt() {
	char c; while ((c = getchar()) < '0' || c > '9'); int r = c - '0';
	while ((c = getchar()) >= '0' && c <= '9') (r *= 10) += c - '0';
	return r;
}

int main() {
	while (n = nextInt()) {
		memset(opt, 0, sizeof opt);
		for (int i = 1; i <= n; ++i)
			p[i] = nextInt();
		sort(p + 1, p + n + 1);
		m = nextInt();
		for (int i = 1; i < n; ++i)
			for (int j = m - 5; j >= p[i]; --j)
				if (opt[j - p[i]] + p[i] > opt[j])
					opt[j] = opt[j - p[i]] + p[i];
		if (m < 5) printf("%d
", m);
		else printf("%d
", m - opt[m - 5] - p[n]);
	}
	return 0;
}
原文地址:https://www.cnblogs.com/johnsonyip/p/5668760.html