题解 CF1025A 【Doggo Recoloring】

水题！

STL大法好！

先来分析一下，可以得知，只要有一种字符出现过两次以上，就可以和别的字符合并，就是Yes，注意字符串长度为1也是Yes

那怎么统计呢？对了，map！！！

代码如下（char[]和unordered_map和string都有哦）:

#include <iostream>
#include <cstring>
#include <unordered_map>
using namespace std;

unordered_map <char, int> mp;

string a;

int b;

int main()
{
    cin >> b >> a;
    int x = a.length();
    for(int i = 0; i <= x - 1; i++)
    {
        mp[a[i]]++;
    }
    for(unordered_map <char, int>::iterator it = mp.begin(); it != mp.end(); ++it)
    {
        if(it -> second >= 2)
        {
            cout << "Yes\n";
            return 0;
        }
    }
    cout << (x == 1 ? "Yes" : "No") << endl;
    return 0;
}

以上是string+unordered_map

#include <iostream>
#include <cstring>
#include <map>
using namespace std;

map <char, int> mp;

string a;

int b;

int main()
{
    cin >> b >> a;
    int x = a.length();
    for(int i = 0; i <= x - 1; i++)
    {
        mp[a[i]]++;
    }
    for(map <char, int>::iterator it = mp.begin(); it != mp.end(); ++it)
    {
        if(it -> second >= 2)
        {
            cout << "Yes\n";
            return 0;
        }
    }
    cout << (x == 1 ? "Yes" : "No") << endl;
    return 0;
}

map+string的

#include <iostream>
#include <cstring>
#include <map>
using namespace std;

map <char, int> mp;

char a[100005];

int b;

int main()
{
    cin >> b >> a;
    int x = strlen(a);
    for(int i = 0; i <= x - 1; i++)
    {
        mp[a[i]]++;
    }
    for(map <char, int>::iterator it = mp.begin(); it != mp.end(); ++it)
    {
        if(it -> second >= 2)
        {
            cout << "Yes\n";
            return 0;
        }
    }
    cout << (x == 1 ? "Yes" : "No") << endl;
    return 0;
}

char+map

#include <iostream>
#include <cstring>
#include <unordered_map>
using namespace std;

unordered_map <char, int> mp;

char a[100005];

int b;

int main()
{
    cin >> b >> a;
    int x = strlen(a);
    for(int i = 0; i <= x - 1; i++)
    {
        mp[a[i]]++;
    }
    for(unordered_map <char, int>::iterator it = mp.begin(); it != mp.end(); ++it)
    {
        if(it -> second >= 2)
        {
            cout << "Yes\n";
            return 0;
        }
    }
    cout << (x == 1 ? "Yes" : "No") << endl;
    return 0;
}