这题我一下子就想到思路了,直接AC:
n列,m个方块,ci表示第i个方块落在第ci列;
只需要建立一个count数组,记录每列所落下的方块数,最后找出count数组中的最小值即可。(短板效应)
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
int n, m;
cin >> n >> m;
vector<int>nums(m);//m个方块
vector<int>count(n + 1, 0);//n列
for (int i = 0; i < m; i++)
{
cin >> nums[i];
count[nums[i]]++;
}
int score = count[1];
for (int i = 2; i <= n; i++)
{
score = min(score, count[i]);
}
cout << score << endl;
system("pause");
return 0;
}