• PAT甲级 1016 Phone Bills (25分)(坑点)


    坑点1: 如果该用户的消费不满足有效条件 那么他的总消费也不必输出

    坑点2:最后一组样例输出总消费时 应该考虑最后一组样例是否满足有效条件

    有效条件: 第i个记录与第i+1条记录的名字相同并且第i个记录的状态是 "on-line" 第i+1个记录的状态是"off-line";

    kswl

     1 #include <cstdio>
     2 #include <cstring>
     3 #include <iostream>
     4 #include <algorithm>
     5 using namespace std;
     6 const int maxn = 0x3f3f3f3f;
     7 typedef long long ll;
     8 int bill[25];
     9 struct bl{
    10     string name;
    11     int mon,day,hour,minu;
    12     string st;// state on-line or off-line
    13 }b[1005];
    14 string a = "on-line";
    15 string c = "off-line";
    16 bool cmp(bl p,bl q) {//排序部分
    17     if(p.name != q.name) return  p.name < q.name;
    18     else if(p.mon != q.mon) return p.mon < q.mon;
    19     else if(p.day != q.day) return p.day < q.day;
    20     else if(p.hour != q.hour) return p.hour < q.hour;
    21     return p.minu < q.minu;
    22 }
    23 int main()
    24 {
    25     for(int i = 0; i < 24; i++) {
    26         cin >> bill[i];
    27     }
    28     int n;
    29     cin >> n;
    30     for(int i = 0; i < n ; i++) {
    31         cin >> b[i].name;
    32         scanf("%d:%d:%d:%d",&b[i].mon,&b[i].day,&b[i].hour,&b[i].minu);
    33         cin >> b[i].st;
    34     }
    35     sort(b,b+n,cmp);
    36     int t = 0;
    37     double total = 0;
    38     bool flag = false;
    39     for(int i = 0; i < n-1; i++) {
    40            double money = 0;
    41             if(b[i].name == b[i+1].name && b[i].st == a && b[i+1].st == c) {
    42                int sum = b[i+1].day*60*24 + b[i+1].hour*60 + b[i+1].minu - (b[i].day*60*24 + b[i].hour*60 + b[i].minu);
    43                for(int p = b[i].day; p <= b[i+1].day; p++) {//一分一分的算消费
    44                 int j = 0,jj = 23;
    45                 if(p == b[i].day) j = b[i].hour;
    46                 if(p == b[i+1].day) jj = b[i+1].hour;
    47                for(; j <= jj; j++) {
    48                    int k = 0,kk = 60;
    49                    if(j == b[i].hour&&p == b[i].day) k = b[i].minu;
    50                    if(j == b[i+1].hour&&p == b[i+1].day) kk = b[i+1].minu;
    51                    for(;k < kk; k++) {
    52                     money += bill[j];
    53                    }
    54                }
    55                }
    56                money = money*1.0/100.0;
    57                total += money;
    58                if(t == 0) {
    59                 cout << b[i].name << " ";
    60                 if(b[i].mon < 10) cout << 0;
    61                 cout << b[i].mon << endl;
    62                 t = 1;
    63                }
    64                printf("%02d:%02d:%02d %02d:%02d:%02d ",b[i].day,b[i].hour,b[i].minu,b[i+1].day,b[i+1].hour,b[i+1].minu);
    65                printf("%d $%.2lf
    ",sum,money);
    66             } else if(b[i].name != b[i+1].name) {
    67                t = 0;
    68                if(total != 0){//坑点1:当total不等于0时同时也说明该用户的消费满足有效条件
    69                printf("Total amount: $%.2lf
    ",total);
    70                }
    71                total = 0;
    72             }
    73     }
    74     if(t == 1)//坑点2:最后一组也要判断是否满足有效条件
    75     printf("Total amount: $%.2lf
    ",total);
    76     return 0;
    77 }
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  • 原文地址:https://www.cnblogs.com/LLLAIH/p/12256269.html
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