--g++||c++

A - An easy problem

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice HDU 2055

Description

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).

 

Input

On the first line, contains a number T.then T lines follow, each line is a case.each case contains a letter and a number.

 

Output

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

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;



int  main(){
	int f[1010];
	int j = 1;
	for(int i = 'A';i<='Z';i++)
	f[i] = j++;
	j = 1;
	for(int i = 'a';i<='z';i++)
	f[i] = -j++;
	int t;
	scanf("%d",&t);
	while(t--){
		int d;
		char a[1];//G++会过，c++不会过，改成2即可。
		scanf("%s%d",a,&d);
		printf("%d\n",d+f[a[0]]);
	}
	
	
}