PAT-GPLT训练集 L1-043 阅览室

PAT-GPLT训练集 L1-043 阅览室

注意：连续的S和E才算一次借还

代码：

#include<iostream>
#include<cstdio>
using namespace std;
typedef struct {
    int id;
    char record;
    int m;
} P;
int main() {
    int n;
    scanf("%d", &n);
    while(n--) {
        P c[10000];
        int id, a, b;
        char record;
        int cur = 0, count = 0;
        double sum = 0;
        while(1) {
            scanf("%d %c %d:%d", &id, &record, &a, &b);
            if(id) {
                c[cur].id = id; c[cur].record = record;
                c[cur].m = a*60 + b;
                cur++;
            } else {
                for(int i = 0; i < cur; i++) {
                    if(c[i].record == ‘S‘) for(int j = i+1; j < cur; j++) {
                        if(c[j].record == ‘S‘ && c[i].id == c[j].id) break;
                        if(c[j].record == ‘E‘ && c[i].id == c[j].id) {
                            sum += c[j].m - c[i].m;
                            count++;
                            break;
                        }
                    }
                }
                if(count) printf("%d %d\n", count, (int)((sum/count)+0.5));
                else printf("0 0\n");
                break;
            }
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/kindleheart/p/8669082.html