PAT A 1016. Phone Bills (25)【模拟】

题目：https://www.patest.cn/contests/pat-a-practise/1016

思路：用结构体存储，按照名字和日期排序，然后先判断是否有效，然后输出，时间加减直接暴力即可

 #include<set>
 #include<map>
 #include<queue>
 #include<algorithm>
 #include<iostream>
 #include<cstdio>
 #include<vector>
 #include<string>
 #include<string.h>
 using namespace std;
 typedef long long LL;
 const int INF = 0x7FFFFFFF;
 const int maxn = 1e3 + ;
 double f[maxn];
 int n, x;
 char ss[maxn];

 struct point
 {
     string s;
     int a, b, c, d, f;
     void read()
     {
         cin >> s;
         scanf("%d:%d:%d:%d", &a, &b, &c, &d);
         scanf("%s", ss);
         if (ss[] == 'n')f = ; else f = ;
     }
     bool operator<(const point &x)const
     {
         return s == x.s ? a == x.a ? b == x.b ? c == x.c ? d < x.d : c < x.c : b < x.b : a < x.a : s < x.s;
     }
     bool operator==(const point &x)const
     {
         return b == x.b&&c == x.c&&d == x.d;
     }
 }a[maxn];

 double putout(point x, point y)
 {
     double ans = ;
     int cnt = ;
     point u = x;
     while (true)
     {
         ans += f[u.c]; u.d += ;
         u.c += u.d / ;    u.d %= ;
         u.b += u.c / ;    u.c %= ;
         cnt++;  if (u == y) break;
     }
     printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2lf\n", x.b, x.c, x.d, y.b, y.c, y.d, cnt, ans);
     return ans;
 }

 int main()
 {
     for (int i = ; i < ; i++)scanf("%d", &x), f[i] = x / 100.0;
     scanf("%d", &n);
     for (int i = ; i < n; i++)a[i].read();
     sort(a, a + n);
     for (int i = , j; i < n; i = j)
     {
         int flag = , y = a[i].f;
         for (j = i + ; a[j].s == a[i].s; j++)
         {
             if (!y&&a[j].f){ flag = ; break; }
             else y = a[j].f;
         }
         if (flag)
         {
             cout << a[i].s;
             printf(" %02d\n", a[i].a);
             point x = a[i];
             double ans = ;
             for (j = i + ; a[j].s == a[i].s; j++)
             {
                 if (!x.f&&a[j].f)ans += putout(x, a[j]);
                 x = a[j];
             }
             printf("Total amount: $%.2lf\n", ans);
         }
     }
     return ;
 }