SSOJ 1192.加密

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题目:

传送门


分析:

直接暴力查看当前字符是否是前面密文对应的原文,通过 c p p cpp c s t r i n g f i n d cstring-find p a s c a l pascal 中的 p o s pos 可以实现


代码:

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring> 
#include<cstdlib>
#include<algorithm>
#include<set>
#include<queue>
#include<vector>
#include<map>
#include<list>
#include<ctime>
#include<iomanip>
#include<string>
#include<bitset>
#include<deque>
#include<set>
#define LL long long
#define ch cheap
#define XJQ (int)%%%
using namespace std;
inline LL read() {
    LL d=0,f=1;char s=getchar();
    while(s<'0'||s>'9'){if(s=='-')f=-1;s=getchar();}
    while(s>='0'&&s<='9'){d=d*10+s-'0';s=getchar();}
    return d*f;
}
int tf[100010];
int main()
{
	string s,z,c;
	int len=0;
	cin>>s>>z;
	cout<<z;
	for(int i=0;i<z.size();i++)
	{
		if(c.find(z[i])+1)
		  tf[c.find(z[i])]=1;
		else 
			c+=s[z[i]-'a'];
	}
	for(int i=0;i<c.size();++i)
	  if(!tf[i]) cout<<c[i];
	return 0;
	//请自行忽略一下内容
	/*	%%% XJQ
		%%% WYC
		%%% XXY
		%%% ZYC
		%%% HSZGB
		%%% BPM
		%%% YMW
		%%% ZZY
		%%% HJW 	*/
}

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