中文题目就不用解释了   就是裸的二维全然背包

d[i][j]表示消耗i忍耐杀j个怪最多可获得的经验  然后就用全然背包来做了  二维背包背包只是是多了一重循环

<span style="font-family:Arial Black;">#include<cstdio>

#include<cstring>

#include<algorithm>

using namespace std;

const int N = 105;

int main()

{

    //还需的经验值n。保留的忍耐度m。怪的种数k,最多的杀怪数s

    int n, m, k, s, d[N][N], exp[N], cos[N];

    while (~scanf ("%d%d%d%d", &n, &m, &k, &s))

    {

        memset (d, 0, sizeof (d));

        for (int i = 1; i <= k; ++i)

            scanf ("%d%d", &exp[i], &cos[i]);

        for (int u = 1; u <= k; ++u)

            for (int i = cos[u]; i <= m; ++i)

                for (int j = 1; j <= s; ++j)

                    d[i][j] = max (d[i][j], d[i - cos[u]][j - 1] + exp[u]);

        int i;

        for (i = 1; i <= m; ++i)

            if (d[i][s] >= n) break;

        if (i <= m) printf ("%d\n", m - i);

        else printf ("-1\n");

    }

    return 0;

}

</span>

FATE

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