标签:
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 }
标签:
原文地址:http://www.cnblogs.com/takeoffyoung/p/4318452.html