HDU 2055 An easy problem
1426 ワード
An easy problem
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 17042 Accepted Submission(s): 11448
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
Sample Output
すいもん
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 17042 Accepted Submission(s): 11448
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
すいもん
</pre><pre name="code" class="cpp">#include <stdio.h>
int fun(char ch)
{
if(ch>='a' && ch<='z')
return -(ch-'a'+1);
else
return ch-'A'+1;
}
int main()
{
int T;
int a;
char ch;
scanf("%d%*c",&T);
while(T--)
{
scanf("%c %d",&ch,&a);
getchar();
printf("%d
",fun(ch)+a);
}
return 0;
}