poj_1276

dp[i] 代表能否凑出总数为i的cash

 1 #include <cstdio>

 2 #include <cstring>

 3 

 4 int n[11], d[11], cash, N;

 5 bool dp[100010];

 6 

 7 int main(int argc, char const *argv[])

 8 {

 9     // freopen("in", "r", stdin);

10     while(~scanf("%d%d", &cash, &N)){

11         for(int i = 1; i <= N; ++i)

12             scanf("%d %d", &n[i], &d[i]);

13         memset(dp, false, sizeof(dp));

14         dp[0] = true;

15         for(int i = 1; i <= N; ++i)

16             for(int j = cash; j >= 0; --j)

17                 if(dp[j] == true)

18                     for(int k = 1; k <= n[i]; ++k){

19                         if( j + k*d[i] > cash )

20                             break;

21                         dp[j + k*d[i]] = true;

22                     }

23         for(int i = cash; i >= 0; --i)

24             if(dp[i] == true){

25                 printf("%d\n", i);

26                 break;

27             }

28     }

29     return 0;

30 }