1117. Eddington Number 解析

其实题目不难，就是很绕，然而我现在还是有点绕在里面的。统计数出现的次数，然后按要求输出。

#include <iostream>#include <cstring>#include <map>#define MAX 1000100using namespace std;map <int, int> no2num;int Count[MAX];int n;int ans;int main() {cin >> n;memset(Count, 0, sizeof(Count));int temp;int max = 0;for (int i = 0; i < n; i++) {scanf("%d", &temp);if (temp > max)max = temp;Count[temp]++;}int p = 0;for (int i = 0; i <= max; i++) {if (Count[i] != 0)no2num[++p] = i;}int sum = 0;int ans = 0;for (int i = p; i > 0; i--) {sum += Count[no2num[i]];//cout << sum << " " << no2num[i] << endl;if (sum >= no2num[i] - 1) {ans = no2num[i] - 1;break;}}cout << ans << endl;return 0;}