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;
}