1550: Simple String （做得少的思维题，两个字符串能否组成另外一个字符串问题）

时间：2024-01-14 12:45:44

1550: Simple String

Submit Page    Summary    Time Limit: 1 Sec     Memory Limit: 256 Mb     Submitted: 682     Solved: 287

Description

Welcome,this is the 2015 3th Multiple Universities Programming Contest ,Changsha ,Hunan Province. In order to let you feel fun, ACgege will give you a simple problem. But is that true? OK, let’s enjoy it.
There are three strings A , B and C. The length of the string A is 2*N, and the length of the string B and C is same to A. You can take N characters from A and take N characters from B. Can you set them to C ?

Input

There are several test cases.
Each test case contains three lines A,B,C. They only contain upper case letter.
0<N<100000
The input will finish with the end of file.

Output

For each the case, if you can get C, please print “YES”. If you cann’t get C, please print “NO”.

Sample Input

AABB

BBCC

AACC

AAAA

BBBB

AAAA

Sample Output

YES

NO

Hint

Source

题目意思：
给你三个字符串
a，b，c
长度都是2*n
问你从a和b中各抽出n给字符
能不能组成c
分析：
统计a，b，c中26个字符出现的次数
如果某字符在a中出现次数+在b中出现次数小于该字符在c中出现次数
那么肯定不能组成c
1.A+B 与C的交集必须>=n
2.A与C的交集必须>=n/2，B与C的交集必须>=n/2。
理解上面两句话就可以了
#include<stdio.h>

#include<iostream>

#include<vector>

#include <cstring>

#include <stack>

#include <cstdio>

#include <cmath>

#include <queue>

#include <algorithm>

#include <vector>

#include <set>

#include <map>

#include<string>

#include<string.h>

#include<math.h>

typedef long long LL;

using namespace std;

#define max_v 125

int c1[];

int c2[];

int c3[];

void init()

{

    memset(c1,,sizeof(c1));

    memset(c2,,sizeof(c2));

    memset(c3,,sizeof(c3));

}

int main()

{

    string str1,str2,str3;

    while(cin>>str1)

    {

        init();

        cin>>str2;

        cin>>str3;

        int l=str1.length();

        for(int i=; i<l; i++)

        {

            c1[str1[i]-'A']++;

            c2[str2[i]-'A']++;

            c3[str3[i]-'A']++;

        }

        int flag=;

        int v1=;

        int v2=;

        for(int i=; i<; i++)

        {

            if(c1[i]+c2[i]<c3[i])

            {

                flag=;

                break;

            }

            v1+=max(,c3[i]-c1[i]);

            v2+=min(c3[i],c2[i]);

        }

        if(v2<l/||v1>l/)

            flag=;

        if(flag)

            printf("YES\n");

        else

            printf("NO\n");

    }

    return ;

}

/*

题目意思：

给你三个字符串

a，b，c

长度都是2*n

问你从a和b中各抽出n给字符

能不能组成c

分析：

统计a，b，c中26个字符出现的次数

如果某字符在a中出现次数+在b中出现次数小于该字符在c中出现次数

那么肯定不能组成c

1.A+B 与C的交集必须>=n

2.A与C的交集必须>=n/2，B与C的交集必须>=n/2。

理解上面两句话就可以了

*/

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