NYOJ-a letter and a number

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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

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

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

样例输入

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

样例输出

19
18
10
-17
-14
-4

代码:

#include<stdio.h> 
int main()
{
	int t,y;
	char ch;
	scanf("%d",&t);
	while(t--)
	{
		getchar();  //注意字符输入时可能出现的问题,否则ch可能会接收'\n';
		scanf("%c%d",&ch,&y);
		if(ch>='a'&&ch<='z')
		printf("%d\n",y-(ch-'a'+1));
		else
		printf("%d\n",y+(ch-'A'+1));
	}
	return 0;
}

NYOJ-a letter and a number

标签：style   color   io   os   ar   for   div   sp   问题   

原文地址：http://blog.csdn.net/qq_18062811/article/details/39431925