C语言一道简单的题目An Easy ProblemTime Limit:1000MS Memory Limit:65536KTotal Submit:16 Accepted:7 Description 定义: f(A)=1 f(a)=-1 f(B)=2 f(b)=-2 . . . . . . f(Z)=26 f(z)=-26 读入一个字符x和一个整数y,请你求出f(x)+yInput 第
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![C语言一道简单的题目An Easy ProblemTime Limit:1000MS Memory Limit:65536KTotal Submit:16 Accepted:7 Description 定义: f(A)=1 f(a)=-1 f(B)=2 f(b)=-2 . . . . . . f(Z)=26 f(z)=-26 读入一个字符x和一个整数y,请你求出f(x)+yInput 第](/uploads/image/z/10131483-3-3.jpg?t=C%E8%AF%AD%E8%A8%80%E4%B8%80%E9%81%93%E7%AE%80%E5%8D%95%E7%9A%84%E9%A2%98%E7%9B%AEAn+Easy+ProblemTime+Limit%3A1000MS++Memory+Limit%3A65536KTotal+Submit%3A16+Accepted%3A7+Description+%E5%AE%9A%E4%B9%89%3A+f%28A%29%3D1+f%28a%29%3D-1+f%28B%29%3D2+f%28b%29%3D-2+.+.+.+.+.+.+f%28Z%29%3D26+f%28z%29%3D-26+%E8%AF%BB%E5%85%A5%E4%B8%80%E4%B8%AA%E5%AD%97%E7%AC%A6x%E5%92%8C%E4%B8%80%E4%B8%AA%E6%95%B4%E6%95%B0y%2C%E8%AF%B7%E4%BD%A0%E6%B1%82%E5%87%BAf%28x%29%2ByInput+%E7%AC%AC)
C语言一道简单的题目An Easy ProblemTime Limit:1000MS Memory Limit:65536KTotal Submit:16 Accepted:7 Description 定义: f(A)=1 f(a)=-1 f(B)=2 f(b)=-2 . . . . . . f(Z)=26 f(z)=-26 读入一个字符x和一个整数y,请你求出f(x)+yInput 第
C语言一道简单的题目
An Easy Problem
Time Limit:1000MS Memory Limit:65536K
Total Submit:16 Accepted:7
Description
定义:
f(A)=1 f(a)=-1
f(B)=2 f(b)=-2
. .
. .
. .
f(Z)=26 f(z)=-26
读入一个字符x和一个整数y,请你求出f(x)+y
Input
第一行为一个整数n,代表有几组测试数据,后面n行,每行一个字符x和一个整数y
Output
一组测试输出一个结果
Sample Input
4
R 1
P 2
G 3
r 1
Sample Output
19
18
10
-15
这是acm答题系统上的题目 各位回答尽量按照要求
C语言一道简单的题目An Easy ProblemTime Limit:1000MS Memory Limit:65536KTotal Submit:16 Accepted:7 Description 定义: f(A)=1 f(a)=-1 f(B)=2 f(b)=-2 . . . . . . f(Z)=26 f(z)=-26 读入一个字符x和一个整数y,请你求出f(x)+yInput 第
#include
int main()
{
int n;
char x;int y;
int sum;
scanf( "%d",&n );
getchar( );
for ( int i = 0; i < n; i++ )
{
scanf( "%c%d",&x,&y );
getchar( );
if ( x > 'Z')
sum = 'a' - x - 1 + y;
else
sum = x - 'A' + 1 + y;
printf( "%d\n",sum );
}
return 0;
}
不过f(r)是-18,故r 1 应该是-17,你的测试数据有问题吧?