天梯赛 L1 阅览室

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/ccDLlyy/article/details/79943619

题目链接：点击打开链接

思路：利用STL map即可

#include <bits/stdc++.h>
using namespace std;
map<int,char> m1;
map<int,int> m2;
int main(){
	int n;
	int cnt = 0;
	double sum = 0;
	int num;
	char op;
	int h,m;
	cin >> n;
	for(int i = 0;i < n;i++){
		m1.clear();
		m2.clear();
		cnt = 0;
		sum = 0.0;
		while(scanf("%d",&num) != EOF){
			getchar();
			op = getchar();
			scanf("%d:%d",&h,&m);
			if(num == 0) break;
			if(op == 'S'){
				m1[num] = op;
				m2[num] = h * 60 + m;
			}
			else if(op == 'E'){
				if(m1.find(num) != m1.end() && m1[num] == 'S'){
					cnt++;
					sum += h*60 + m - m2[num];
					m1[num] = '0';
				}
			}
		}
		if(cnt == 0){
			printf("0 0\n");
		}
		else{
			printf("%d %d\n",cnt,(int)(sum / cnt + 0.5));
		}
	}
	return 0;
}