杭电oj An easy problem

</pre><h1 style="color: rgb(26, 92, 200);">An easy problem</h1><strong><span style="color: green; font-family: Arial;">Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 14028    Accepted Submission(s): 9462</span></strong><div align="left" class="panel_title">Problem Description</div><div class="panel_content">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).</div><div class="panel_bottom"> </div><div align="left" class="panel_title">Input</div><div class="panel_content">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.</div><div class="panel_bottom"> </div><div align="left" class="panel_title">Output</div><div class="panel_content">for each case, you should the result of y+f(x) on a line.</div><div class="panel_bottom"> </div><div align="left" class="panel_title">Sample Input</div><div class="panel_content"><pre><div style="font-family: Courier New,Courier,monospace;">6R 1P 2G 3r 1p 2g 3</div>

 

Sample Output

191810-17-14-4

 

Author

8600

 

Source

校庆杯Warm Up

 

我的代码：//可能不是最优。。也可能很简单。。先判断输入的字符的大写的还是小写的，然后对应的数和输入的数字相加。。

#include<iostream>#include<stdio.h>using namespace std;int main(){int T,i;char x;double y,sum;cin>>T;while(T--){cin>>x>>y;if(x>='A'&&x<='Z'){for(i=0;i<26;i++){if(x=='A'+i){sum=i+1+y;     break;}}}if(x>='a'&&x<='z'){for(i=0;i<26;i++){if(x=='a'+i){sum=-i-1+y;     break;}}}printf("%.0lf\n",sum);}return 0;}