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;
}