nyoj 217-a letter and a number (char)

217-a letter and a number

内存限制:64MB 时间限制:3000ms 特判: No
通过数:4 提交数:5 难度:1

题目描述:

we define f(A) = 1, f(a) = -1, f(B) = 2, f(b) = -2, ... f(Z) = 26, f(z) = -26;
Give you a letter x and a number y , you should output the result of y+f(x).

输入描述:

On the first line, contains a number T(0<T<=10000).then T lines follow, each line is a case.each case contains a letter x and a number y(0<=y<1000).

输出描述:

for each case, you should the result of y+f(x) on a line

样例输入:

复制

6
R 1
P 2
G 3
r 1
p 2
g 3

样例输出:

19
18
10
-17
-14
-4

C/C++ AC：

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <stack>
#include <set>
#include <map>
#include <queue>
#include <climits>
#define PI 3.1415926

using namespace std;
int n;

int main()
{
    cin >>n;
    while (n --)
    {
        char letter;
        int number;
        cin >>letter >>number;
        if (letter >= 'A' && letter <= 'Z')
        {
            cout <<letter - 'A' + 1 + number <<endl;
        }
        else
        {
            cout <<-1 * (letter - 'a' + 1) + number <<endl;
        }
    }
}