POJ 1298 The Hardest Problem Ever 简单暴力

本文介绍了一种经典的凯撒密码解密方法,并通过具体的样例输入和输出展示了如何实现解密过程。通过对密文字符进行偏移还原,成功地揭示了原始消息的内容。

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C - The Hardest Problem Ever
Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%lld & %llu

Description

Julius Caesar lived in a time of danger and intrigue. The hardest situation Caesar ever faced was keeping himself alive. In order for him to survive, he decided to create one of the first ciphers. This cipher was so incredibly sound, that no one could figure it out without knowing how it worked. 

You are a sub captain of Caesar's army. It is your job to decipher the messages sent by Caesar and provide to your general. The code is simple. For each letter in a plaintext message, you shift it five places to the right to create the secure message (i.e., if the letter is 'A', the cipher text would be 'F'). Since you are creating plain text out of Caesar's messages, you will do the opposite: 


Cipher text 
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 

Plain text 
V W X Y Z A B C D E F G H I J K L M N O P Q R S T U 

Only letters are shifted in this cipher. Any non-alphabetical character should remain the same, and all alphabetical characters will be upper case.

Input

Input to this problem will consist of a (non-empty) series of up to 100 data sets. Each data set will be formatted according to the following description, and there will be no blank lines separating data sets. All characters will be uppercase. 


A single data set has 3 components: 

  1. Start line - A single line, "START" 
  2. Cipher message - A single line containing from one to two hundred characters, inclusive, comprising a single message from Caesar 
  3. End line - A single line, "END" 

Following the final data set will be a single line, "ENDOFINPUT". 

Output

For each data set, there will be exactly one line of output. This is the original message by Caesar.

Sample Input

START
NS BFW, JAJSYX TK NRUTWYFSHJ FWJ YMJ WJXZQY TK YWNANFQ HFZXJX
END
START
N BTZQI WFYMJW GJ KNWXY NS F QNYYQJ NGJWNFS ANQQFLJ YMFS XJHTSI NS WTRJ
END
START
IFSLJW PSTBX KZQQ BJQQ YMFY HFJXFW NX RTWJ IFSLJWTZX YMFS MJ
END
ENDOFINPUT

Sample Output

IN WAR, EVENTS OF IMPORTANCE ARE THE RESULT OF TRIVIAL CAUSES
I WOULD RATHER BE FIRST IN A LITTLE IBERIAN VILLAGE THAN SECOND IN ROME
DANGER KNOWS FULL WELL THAT CAESAR IS MORE DANGEROUS THAN HE

循环+暴力

#include <iostream> 
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
const int inf_int = 2e9;
const long long inf_ll = 2e18;
#define inf_add 0x3f3f3f3f
#define mod 1000000007
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=5e2+10;
using namespace std;
typedef  long long ll;
typedef  unsigned long long  ull; 
inline int read(){int ra,fh;char rx;rx=getchar(),ra=0,fh=1;
while((rx<'0'||rx>'9')&&rx!='-')rx=getchar();if(rx=='-')
fh=-1,rx=getchar();while(rx>='0'&&rx<='9')ra*=10,ra+=rx-48,
rx=getchar();return ra*fh;}
//#pragma comment(linker, "/STACK:102400000,102400000")
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}

char s[100000];
int flag = 0;
int main()
{
	while(gets(s)!=NULL)
	{
		//puts(s);
		if(strcmp(s,"ENDOFINPUT")==0) 
		{
			break;
		} 
		if(strcmp(s,"START")==0)
		{
			flag = 1;
			continue;
		}
		if(strcmp(s,"END")==0) 
		{
			flag=0; 
			continue; 
		} 
	
		if(flag==1)
		{
			for(int i=0;i<strlen(s);i++)
			{
				if(s[i]>='A'&&s[i]<='Z')
				{
					if(s[i]<'F')
					{
						s[i] = 'Z' -(4-(s[i]-'A')) ;
						continue;
					}
					s[i] = (s[i]-'A'-5)+'A';	
				} 
			}
			puts(s);
			continue;
		}
		
	
		
	}
	
	return 0;
} 




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