题目:http://acm.hdu.edu.cn/showproblem.php?pid=2055
An easy problem
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 21202 Accepted Submission(s):
14050
Problem 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).
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.
Sample Input
6
R 1
P 2
G 3
r 1
p 2
g 3
Sample Output
19
18
10
-17
-14
-4
Author
8600
思路:用字符串输入后,把字母转化成对应的数字,再相加
难度:这个也是简单的
代码:
1 #include<stdio.h> 2 int main() 3 { 4 int n,i,a; 5 char s; 6 while(scanf("%d",&n)!=EOF) 7 { 8 getchar(); 9 for(i=0;i<n;i++) 10 { 11 scanf("%c%d",&s,&a); 12 getchar(); 13 if(s>='A'&&s<='Z') 14 printf("%d ",s-64+a); 15 else printf("%d ",96-s+a); 16 } 17 } 18 return 0; 19 }