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）

阅读目录

• Description
• Input
• Output
• Sample Input
• Sample Output
• Hint
• Source

 

1550: Simple String

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

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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 ?

 

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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.

 

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Output

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

 

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Sample Input

AABB
BBCC
AACC
AAAA
BBBB
AAAA

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Sample Output

YES
NO

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Hint

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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[30];
int c2[30];
int c3[30];
void init()
{
    memset(c1,0,sizeof(c1));
    memset(c2,0,sizeof(c2));
    memset(c3,0,sizeof(c3));
}
int main()
{
    string str1,str2,str3;
    while(cin>>str1)
    {
        init();
        cin>>str2;
        cin>>str3;
        int l=str1.length();
        for(int i=0; i<l; i++)
        {
            c1[str1[i]-'A']++;
            c2[str2[i]-'A']++;
            c3[str3[i]-'A']++;
        }
        int flag=1;
        int v1=0;
        int v2=0;
        for(int i=0; i<26; i++)
        {
            if(c1[i]+c2[i]<c3[i])
            {
                flag=0;
                break;
            }
            v1+=max(0,c3[i]-c1[i]);
            v2+=min(c3[i],c2[i]);
        }
        if(v2<l/2||v1>l/2)
            flag=0;
        if(flag)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}
/*
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

*/