Codeforces Round #619 (Div. 2)：A. Three Strings

秒速五厘米 2023-07-05 11:40 32阅读 0赞

**Discription**  
You are given three strings a, b and c of the same length n. The strings consist of lowercase English letters only. The i-th letter of a is ai, the i-th letter of b is bi, the i-th letter of c is ci.

For every i (1≤i≤n) you must swap (i.e. exchange) ci with either ai or bi. So in total you’ll perform exactly n swap operations, each of them either ci↔ai or ci↔bi (i iterates over all integers between 1 and n, inclusive).

For example, if a is “code”, b is “true”, and c is “help”, you can make c equal to “crue” taking the 1-st and the 4-th letters from a and the others from b. In this way a becomes “hodp” and b becomes “tele”.

Is it possible that after these swaps the string a becomes exactly the same as the string b?

**Input**  
The input consists of multiple test cases. The first line contains a single integer t (1≤t≤100) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a string of lowercase English letters a.

The second line of each test case contains a string of lowercase English letters b.

The third line of each test case contains a string of lowercase English letters c.

It is guaranteed that in each test case these three strings are non-empty and have the same length, which is not exceeding 100.

**Output**  
Print t lines with answers for all test cases. For each test case:

If it is possible to make string a equal to string b print “YES” (without quotes), otherwise print “NO” (without quotes).

You can print either lowercase or uppercase letters in the answers.

**Example**  
**input**

4
    aaa
    bbb
    ccc
    abc
    bca
    bca
    aabb
    bbaa
    baba
    imi
    mii
    iim

**output**

NO
    YES
    YES
    NO

**Note**  
In the first test case, it is impossible to do the swaps so that string a becomes exactly the same as string b.

In the second test case, you should swap ci with ai for all possible i. After the swaps a becomes “bca”, b becomes “bca” and c becomes “abc”. Here the strings a and b are equal.

In the third test case, you should swap c1 with a1, c2 with b2, c3 with b3 and c4 with a4. Then string a becomes “baba”, string b becomes “baba” and string c becomes “abab”. Here the strings a and b are equal.

In the fourth test case, it is impossible to do the swaps so that string a becomes exactly the same as string b.

**题意**  
给定三个长度相等的字符串a,b,c，要求执行n次某操作：将ci与ai或者bi进行交换，必须执行。  
问n次操作后能否得到a串等于b串。

**思路**  
直接分析若在第i位上，ai,bi,ci都不行等则一定不成立，若，ai,bi相同，但是ci与ai和bi都不相同，则也不成立。

**AC代码**

#include<bits/stdc++.h>
    using namespace std;
    typedef long long ll;
    #define sd(n) scanf("%d", &n)
    #define sdd(n, m) scanf("%d%d", &n, &m)
    #define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
    #define sld(n) scanf("%lld", &n)
    #define sldd(n, m) scanf("%lld%lld", &n, &m)
    #define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
    #define sf(n) scanf("%lf", &n)
    #define sff(n, m) scanf("%lf%lf", &n, &m)
    #define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
    #define pd(n) printf("%d\n", (n))
    #define pdd(n, m) printf("%d %d\n", n, m)
    #define pld(n) printf("%lld\n", n)
    #define pldd(n, m) printf("%lld %lld\n", n, m)
    #define rep(i, a, n) for(int i=a; i<=n; i++)
    #define per(i, n, a) for(int i=n; i>=a; i--)
     
    int T;
    string a,b,c;
    int f;
     
    int main()
    { 
        scanf("%d",&T);
        while(T--)
        { 
            cin>>a>>b>>c;
            f=0;
            for(int i=0;i<a.size();i++)
            { 
                if(a[i]!=b[i]&&a[i]!=c[i]&&b[i]!=c[i])
                { 
                    f=1;
                    break;
                }
                else if(a[i]==b[i]&&c[i]!=a[i]&&c[i]!=b[i])
                { 
                    f=1;
                    break;
                }
            }
            if(f==1)
                cout<<"NO"<<endl;
            else
                cout<<"YES"<<endl;
        }
        return 0;
    }