a letter and a number
时间限制：3000 ms | 内存限制：65535 KB
难度：1
- 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
#include<stdio.h> int main(){ int n; scanf("%d",&n); getchar(); int a; char b; while(n--){ int sum=0; scanf("%c %d",&b,&a); if(b>='A'&&b<='Z'){ sum=b-'A'+1+a; printf("%d\n",sum); getchar(); } if(b>='a'&&b<='z'){ sum=-b+96+a; printf("%d\n",sum); getchar(); } } return 0; }