Codeforces Contest 1100 problem B Build a Contest——Map.erase的用法

Arkady coordinates rounds on some not really famous competitive programming platform. Each round features n problems of distinct difficulty, the difficulties are numbered from 1 to n.

To hold a round Arkady needs n new (not used previously) problems, one for each difficulty. As for now, Arkady creates all the problems himself, but unfortunately, he can’t just create a problem of a desired difficulty. Instead, when he creates a problem, he evaluates its difficulty from 1 to n and puts it into the problems pool.

At each moment when Arkady can choose a set of n new problems of distinct difficulties from the pool, he holds a round with these problems and removes them from the pool. Arkady always creates one problem at a time, so if he can hold a round after creating a problem, he immediately does it.

You are given a sequence of problems’ difficulties in the order Arkady created them. For each problem, determine whether Arkady held the round right after creating this problem, or not. Initially the problems pool is empty.

Input
The first line contains two integers n and m (1≤n,m≤105) — the number of difficulty levels and the number of problems Arkady created.

The second line contains m integers a1,a2,…,am (1≤ai≤n) — the problems’ difficulties in the order Arkady created them.

Output
Print a line containing m digits. The i-th digit should be 1 if Arkady held the round after creation of the i-th problem, and 0 otherwise.

Examples
inputCopy
3 11
2 3 1 2 2 2 3 2 2 3 1
outputCopy
00100000001
inputCopy
4 8
4 1 3 3 2 3 3 3
outputCopy
00001000
Note
In the first example Arkady held the round after the first three problems, because they are of distinct difficulties, and then only after the last problem.

题意：

有m道题目，每道题目有一个难度，在1到n之间，从头开始遍历，如果到一个位置，累积的题目的难度是一轮，就是1到n之间所有难度都有，输出1，这些题目删除，否则输出0

题解：

用一个Map计数就好了，删除的时候，不能直接在for里面it++，需要map.erase(it++)。

#include<bits/stdc++.h>
using namespace std;
map<int,int>vis;
int main()
{
    int n,m,a;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
    {
        scanf("%d",&a);
        vis[a]++;
        if(vis.size()==n)
        {
            for(map<int,int>::iterator it=vis.begin();it!=vis.end();it++)
            {
                it->second--;
                if(it->second==0)
                    vis.erase(it);
            }
            printf("1");
        }
        else
            printf("0");
    }
    printf("\n");
    return 0;
}