CodeForces Gym-101350M

M. Make Cents?

time limit per test

6.0 s

memory limit per test

256 MB

input

standard input

output

standard output

Every year, an elephant qualifies to the Arab Collegiate Programming Competition. He graduated this year, but that’s irrelephant. What’s important is that the location of the competition might not have been the same every year. Therefore, after every trip, he always has leftover money in the currency of the country he visited.

Now he wants to see how much Jordanian Dinars he has after all those competitions. Can you help him convert the leftover money from all competitions to Jordanian Dinar, if that makes any cents?

Input

The first line of input is T – the number of test cases.

The first line of each test case contains C and N (1 ≤ C, N ≤ 100000), the number of currency types and the number of competitions, respectively.

The next C lines each contain the name of the currency C_i of maximum length 10 in lowercase and/or uppercase letters, and the value V_iof that currency in Jordanian Dinar (0 < V_i ≤ 1000). The names are case-sensitive.

The next N lines each contains an amount left over from each competition (0 ≤ N_i ≤ 1000), and the name of the currency of that amount (it is guaranteed that the name was either given in the input or is “JD”).

Output

For each test case, print on a single line the total amount of money he has in Jordanian Dinar(JD) rounded to 6 decimal digits.

Example

input

Copy

1
3 5
dollar 0.71
euro 0.76
turkish 0.17
5.1 dollar
6 dollar
7 turkish
3 euro
1.1 JD

output

Copy

12.451000

做这题就会深深地体会到STL的map容器灰常的好用了。题目很简单，掌握map用法就会做了；

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <map>
#include <string>
#include <algorithm>
#define maxn 100010
using namespace std;
int main(){
    int t;
    map<string, double> mp;
    int c, n;
    double sum;
    string str1;
    string str2;
    double v1, v2;
    cin >> t;
    while (t--){
        cin >> c >> n;
        for (int i = 0; i < c; i++){
            cin>>str1;
            cin >> v1;
            mp[str1] = v1;
        }
        mp["JD"] = 1;
        sum = 0;
        for (int i = 0; i < n; i++){
            cin >> v2;
            cin>>str2;
            sum += mp[str2] * v2;
        }
        printf("%.6f
", sum);
    }
    return 0;
}

插入一个我觉得写的挺好的关于map容器的详细讲解链接，最好自己写个代码都用一用就熟练了。

https://www.cnblogs.com/Braveliu/p/6427050.html