PTA (Basic Level) 1039. In the end buying

I want to buy some red beads made a bunch of their favorite beads. There are many stall selling beads strings of colorful beads, but refused to sell any string split up. So red to help you determine what, a beaded yard contains all the beads you want? If so, tell her how much extra beads; if not, then tell her how much missing beads.

For your convenience, we use [0-9], [az], the characters within [AZ] to represent the range of colors. For example, in FIG. 1, the third string is wanted red beads; a first string may then buy, because it contains all the beads she wants, does not require more than the beads 8; the second string can not buy because there is no black beads, and a less red beads.

Input formats:

Each input comprises a test. Each test case separately in two lines has given beads and red beads wanted to stall, no more than 1000 two strings of beads.

Output formats:

If you can buy the output in a row Yesand how much extra beads; if not buy, the output in a row Noas well as lack of the number of beads. Meanwhile separated by a space.

Sample Input 1:

ppRYYGrrYBR2258
YrR8RrY

Output Sample 1:

Yes 8

Sample Input 2:

ppRYYGrrYB225
YrR8RrY

Output Sample 2:

No 2

Thinking

Use mapcontainers to count the number of maps, each test has beads, if one can not find any to illustrate the lack of, or a reduction on the corresponding beads
A1092 ( https://pintia.cn/problem-sets/994805342720868352/problems/994805374509498368 ) is the English version of this title

Code

#include<bits/stdc++.h>
using namespace std;
int main()
{
    map<char,int> mp;
    string a;
    string b;
    getline(cin, a);
    getline(cin, b);
    for(int i=0;i<a.size();i++)
    {
        if(mp.count(a[i]) == 0)
        {
            mp[a[i]] = 1;
        }else mp[a[i]] += 1;
    }
    bool can_make = true;
    int need = 0;
    for(int i=0;i<b.size();i++)
    {
        if(mp.count(b[i]) == 0 || mp[b[i]] == 0)  //找不到指定颜色的珠子或者该珠子用完
        {
            can_make = false;
            need++;
        }else
        {
            mp[b[i]] -= 1;
        }
    }
    if(can_make)
    {
        map<char,int>::iterator it;
        int extra = 0;
        for(it=mp.begin();it!=mp.end();it++)
            extra += it->second;
        cout << "Yes " << extra;
    }else cout << "No " << need;
    return 0;
}

Quote

https://pintia.cn/problem-sets/994805260223102976/problems/994805283241443328