HDU2055An easy problem
1312 ワード
An easy problem
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5908 Accepted Submission(s): 4106
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).
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
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5908 Accepted Submission(s): 4106
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).
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
#include <stdio.h>
// : , getchar()
int main()
{
int t,y, ans;
char x;
scanf("%d", &t);
while(t--)
{
getchar();
ans = 0;
scanf("%c%d", &x, &y);
if(x >= 'a' && x <= 'z')
{
ans = y - (x - 'a' + 1);
}else
{
ans = y + (x - 'A' + 1);
}
printf("%d
", ans);
}
return 0;
}