题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1114
分析：每个物品可以取无限个，完全背包问题，要求装满，所以dp[0]=0,其余为INF；
1 #include<iostream> 2 #include<sstream> 3 #include<cstdio> 4 #include<cstdlib> 5 #include<string> 6 #include<cstring> 7 #include<algorithm> 8 #include<functional> 9 #include<iomanip> 10 #include<numeric> 11 #include<cmath> 12 #include<queue> 13 #include<vector> 14 #include<set> 15 #include<cctype> 16 #define PI acos(-1.0) 17 const int INF = 0x3f3f3f3f; 18 const int NINF = -INF - 1; 19 const int maxn = 1e4 + 5; 20 typedef long long ll; 21 using namespace std; 22 int cost[505], weight[505]; 23 int dp[maxn]; 24 int main() 25 { 26 int T; 27 scanf("%d", &T); 28 while (T--) 29 { 30 int E, F; 31 scanf("%d%d", &E, &F); 32 int n; 33 scanf("%d", &n); 34 for (int i = 0; i < n; ++i) 35 scanf("%d%d", &cost[i], &weight[i]); 36 int ans = F - E; 37 for (int i = 0; i <= ans; ++i) 38 dp[i] = INF; 39 dp[0] = 0; 40 for (int i = 0; i < n; ++i) 41 { 42 for (int j = weight[i]; j <= ans; ++j) 43 { 44 dp[j] = min(dp[j], dp[j - weight[i]] + cost[i]); 45 //cout << j << ' ' << dp[j] << endl; 46 } 47 } 48 if (dp[ans] == INF) printf("This is impossible. "); 49 else printf("The minimum amount of money in the piggy-bank is %d. ", dp[ans]); 50 } 51 return 0; 52 }