标签:ide sed tee mem long into als HERE time
Input
The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers E and F. They indicate the weight of an empty pig and of the pig filled with coins. Both weights are given in grams. No pig will weigh more than 10 kg, that means 1 <= E <= F <= 10000. On the second line of each test case, there is an integer number N (1 <= N <= 500) that gives the number of various coins used in the given currency. Following this are exactly N lines, each specifying one coin type. These lines contain two integers each, Pand W (1 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it‘s weight in grams.
Output
Print exactly one line of output for each test case. The line must contain the sentence "The minimum amount of money in the piggy-bank is X." where X is the minimum amount of money that can be achieved using coins with the given total weight. If the weight cannot be reached exactly, print a line "This is impossible.".
Sample Input
3 10 110 2 1 1 30 50 10 110 2 1 1 50 30 1 6 2 10 3 20 4
Sample Output
The minimum amount of money in the piggy-bank is 60. The minimum amount of money in the piggy-bank is 100. This is impossible.
题目大意:
给定正整数存钱罐的重量E,以及装满后的重量F,然后给n对数据分别代表价值和重量,求装满存钱罐最少的价值。
完全背包模板题
#include <iostream> #include <cstring> using namespace std; const int INF=0x3f3f3f3f; int dp[10005],v[505],w[505]; int main() { ios::sync_with_stdio(false); int T,x,y,n; cin>>T; while(T--) { memset(dp,INF,sizeof dp); cin>>x>>y>>n; y-=x; for(int i=1;i<=n;i++) cin>>v[i]>>w[i]; dp[0]=0; for(int i=1;i<=n;i++) for(int j=w[i];j<=y;j++) dp[j]=min(dp[j],dp[j-w[i]]+v[i]); if(dp[y]!=INF) cout<<"The minimum amount of money in the piggy-bank is "<<dp[y]<<".\n"; else cout<<"This is impossible.\n"; } return 0; }
标签:ide sed tee mem long into als HERE time
原文地址:https://www.cnblogs.com/zdragon1104/p/9188924.html