2159

d[i][j] : 用i的忍耐度，j的杀怪次数，可以获得的最大经验值   状态转移：d[i][j] = max{d[i - a[k]][j - 1] + b[k], 1 <= k <= K} 边界：全部初始化为0 答案：寻找最小的i使得d[i][j] >= N AC:---------------------------------------------------------------------------#include<stdio.h>#include<memory.h>int a[125], b[125], d[105][105];int main(){ //freopen("E:/in.txt", "r", stdin); /// int N, M, K, S, i, j, k; while(scanf("%d%d%d%d", &N, &M, &K, &S) != EOF) {  for(i = 1; i <= K; i++) scanf("%d%d", &a[i], &b[i]);  memset(d, 0, sizeof(d));  for(i = 1; i <= M; i++)  {   for(j = 1; j <= S; j++)    for(k = 1; k <= K; k++) if(i - b[k] >= 0)    {     if(d[i - b[k]][j - 1] + a[k] >= d[i][j])      d[i][j] = d[i - b[k]][j - 1] + a[k];    }    if (d[i][S] >= N) break;  }     printf("%d\n", M - i);   }  return 0;}