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WIKIOI-1277 生活大爆炸

題目描述 Description

Sheldon and Leonard are physicists who are ?xated 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.

謝耳朵和萊納德是研究BB理論的物理學家。要用暗號（大寫字母）聯係。


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.

（讀懂這段話是解題的關鍵，翻譯了就沒意義了）


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.

他們有一個密鑰K。第P個字母有S = 3P + K。


For example, here is how ZOOM is encoded when K = 3. The ?rst 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.

輸入描述 Input Description
The input will be two lines. The ?rst 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.

輸入有兩行。第一行一個正整數K（〈10）。第二行是最多20個大寫字母組成的信。

輸出描述 Output Description
The output will be the decoded word of uppercase letters.

輸入就是解密後的信。

樣例輸入 Sample Input
樣例1：

3

FXAB

樣例2:

5

JTUSUKG

樣例輸出 Sample Output
樣例1：

ZOOM

樣例2：

BIGBANG

 

 

#include<stdio.h>
#include<string.h>
char a[27]={'0','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
char b[2000];
int main()
{
    int i,j,n,m,k,s,flag,x;
    scanf("%d",&n);
    getchar();
    scanf("%s",b);
    m=strlen(b); 
    k=0;
    for(i=0;i<m;i++)
    {
       for(j=1;j<=26;j++)
       {
         if(a[j]==b[i])
         {
           flag=j;
           break;
         }
       }
       s=3*(i+1)+n;
       x=0;
       if(flag-s>0)
       j=flag-s;
       else
       {
           j=flag-s+26;
       }
       b[k++]=a[j];
    }
    for(i=0;i<k;i++)
    printf("%c",b[i]);
    puts("");
    return 0;
}

最後更新：2017-04-03 12:55:26

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