HDU 2055 - An easy problem
阿新 • • 發佈:2018-11-30
An easy problem
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 30364 Accepted Submission(s): 19812
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
直接利用ASCII值就行了。
程式碼:
#include<stdio.h> int main() { int a[70],b[70],i,j,t,sum,m,n; char ch; scanf("%d",&t); while(t--) { j=1; sum=0; for(i=1;i<=26;i++) a[i]=i; for(i=1;i<=26;i++) b[i]=-i; getchar(); scanf("%c",&ch); if(ch>='A'&&ch<='Z') { m=a[ch-'@']; scanf("%d",&n); sum=m+n; printf("%d\n",sum); } else if(ch>='a'&&ch<='z') { m=b[ch-'`']; scanf("%d",&n); sum=m+n; printf("%d\n",sum); } } return 0; }