杭电子科技大学_god ls a girl神翻译

杭电子科技大学_god ls a girl神翻译godisagirlTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:32768/32768K(Java/Others)TotalSubmission(s):1337    AcceptedSubmission(s):618ProblemDescriptionOneday,whenIwa

大家好,又见面了,我是你们的朋友全栈君。如果您正在找激活码,请点击查看最新教程,关注关注公众号 “全栈程序员社区” 获取激活教程,可能之前旧版本教程已经失效.最新Idea2022.1教程亲测有效,一键激活。

Jetbrains全系列IDE稳定放心使用

god is a girl

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1337    Accepted Submission(s): 618

Problem Description
One day,when I was dreaming,god went into my dream,she was pretty cute,just like /^_^\…

I really wanted to talked to her,but my English was so poor and she was not a national god but a foreign one…After thirty minutes,she flew away…but story was not finished here,she had left a letter for me!!!What puzzled me so much is the letter was encoded.I had thought for many days,but still can’t get it. Now I turn to you for help,with some limited prompts,can you help me to decode the whole letter?

Prompts:

GDJIJ,EL SSJT UT YWOSQNIVZMI. -> HELLO,MY NAME IS LINDAINVERS.

CN WLP JRVMFGQ BVR,IJCFI? -> DO YOU REQUIRE AID,HUMAN?

NMAB VYNNF, FI’E VC HP IXJ ZLQZI. -> ONCE AGAIN, IT’S UP TO THE ELVES.



 

Input
There is multy cases,please process to EOF.

Each case is one line of string with uppercase letters and white spaces and other symbols.

 

Output
One line of the decoded string.
 

Sample Input
  
  
  
SGC CGGJX GC BMHVQ BGU BCIHNYNBX GNPLV!
 

Sample Output
  
  
  
THE FLOWS OF MAGIC ARE WHIMSICAL TODAY!
 

Author
Teddy
 

Source
 

Recommend
lcy   |   We have carefully selected several similar problems for you:  
2673 
2668 
2669 
2671 
2670 
 
#include<stdio.h>
#include<string.h>
int main()
{
	char str1[100000],str2[100000];
	int a[100000],i,t,j;
	memset(str1,0,sizeof(str1));
	while(gets(str1))
	{
     	memset(a,0,sizeof(a));         清零应放在while内
	memset(str2,0,sizeof(str2));
	t=strlen(str1);
	a[0]=1;
	a[1]=1;
	for(i=2;i<t;i++)
	  {
	  a[i]=a[i-1]+a[i-2];
    	while(a[i]>26)
	  a[i]-=26;
      }
     for(i=0,j=0;i<t;i++)
	 {
	 	if(str1[i]>='A'&&str1[i]<='Z')
	 	{
	 	str2[i]=str1[i]+a[j++];
	    if(str2[i]>'Z')
	 	str2[i]-=26;
	    }
	    else
	    str2[i]=str1[i];
		 }	
		 puts(str2);
      }
      return 0;
  }
<pre name="code" class="html">#include<stdio.h>
#include<string.h>
int main()
{
	char str1[100000] ;
	int a[100000],i,t,j;
	while(gets(str1))
	{
	t=strlen(str1);
	a[0]=1;
	a[1]=1;
	for(i=2;i<t;i++)
	  {
	  a[i]=a[i-1]+a[i-2];
    	while(a[i]>26)
	  a[i]-=26;
      }
     for(i=0,j=0;i<t;i++)
	 {
	 	if(str1[i]>='A'&&str1[i]<='Z')
	 	{
	 	str1[i]+=a[j++];
	    if(str1[i]>'Z')
	 	str1[i]-=26;
	    }
		 }	
		 puts(str1);
      }
      return 0;
  }



 
版权声明:本文内容由互联网用户自发贡献,该文观点仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请联系我们举报,一经查实,本站将立刻删除。

发布者:全栈程序员-站长,转载请注明出处:https://javaforall.net/184519.html原文链接:https://javaforall.net

(0)
全栈程序员-站长的头像全栈程序员-站长


相关推荐

发表回复

您的邮箱地址不会被公开。 必填项已用 * 标注

关注全栈程序员社区公众号