codevs 1277 生活大爆炸 2012年CCC加拿大高中生信息学奥赛

 时间限制: 1 s

 空间限制: 128000 KB

 题目等级 : 白银 Silver

题目描述 Description

Sheldon and Leonard are physicists who are fixated on the BIG BANG theory. In order to exchange secret insights they have devised a code that encodes UPPERCASE words by shifting their letters forward.

谢耳朵和莱纳德是研究BB理论的物理学家。要用暗号（大写字母）联系。

Shifting a letter by S positions means to go forward S letters in the alphabet. For example, shifting B by S = 3 positions gives E. However, sometimes this makes us go past Z, the last letter of the alphabet. Whenever this happens we wrap around, treating A as the letter that follows Z. For example, shifting Z by S = 2 positions gives B.

（读懂这段话是解题的关键，翻译了就没意义了）

Sheldon and Leonard’s code depends on a parameter K and also varies depending on the position of each letter in the word. For the letter at position P, they use the shift value of S = 3P + K.

他们有一个密钥K。第P个字母有S = 3P + K。

For example, here is how ZOOM is encoded when K = 3. The first letter Z has a shift value of S = 3 × 1 + 3 = 6; it wraps around and becomes the letter F. The second letter, O, has S = 3 × 2 + 3 = 9 and becomes X. The last two letters become A and B. So Sheldon sends Leonard the secret message: FXAB
Write a program for Leonard that will decode messages sent by Sheldon.

输入描述 Input Description

The input will be two lines. The first line will contain the positive integer K (K < 10), which is used to compute the shift value. The second line of input will be the word, which will be a sequence of uppercase characters of length at most 20.

输入有两行。第一行一个正整数K（〈10）。第二行是最多20个大写字母组成的信。

输出描述 Output Description

The output will be the decoded word of uppercase letters.

输入就是解密后的信。

样例输入 Sample Input

样例1：

3

FXAB

样例2:

5

JTUSUKG

样例输出 Sample Output

样例1：

ZOOM

样例2：

BIGBANG

数据范围及提示 Data Size & Hint

 

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>

using namespace std;

int main()
{
    char b[25];
    int i,k,s;
    scanf("%d",&k);
    scanf("%s",b);
    for(i=0;b[i];i++)
    {
        s=3*(i+1)+k;
        if(b[i]-s<'A')
        {
            s-=b[i]-'A';
            b[i]='Z'+1;
        }
        printf("%c",b[i]-s);
    }
}