题目 A1016 Phone Bills
-
题意
给出一天24小时每个小时的分钟计费,然后给出用户及其通话记录,判断其中有效的记录,计算产生费用,最终采用名字升序的顺序给出用户的月账单。 -
思路
主要要实现三个模块:
① 将按题意记录排序(name, day, hour, minute递增)
② 获取其中的有效记录(采用needPrint
去记录,遇到on = 1, 下一个再遇到off = 2 即为有效记录)
③ 计算分钟数及话费 (因为计费是按分钟收费,所以以每分钟的增幅去向off靠拢即可)
注意:
on和off相邻才算是有效的一条通话记录,即x[i] == "on-line" && x[i-1] == "off-line"
。 -
Code in C++
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#define maxn 1001
struct record {
std::string name;
int month;
int day;
int hour;
int minute;
std::string word;
void print() {
std::cout << name << " " << month << ":" << day << ":" << hour << ":" << minute << " " << word << std::endl;
}
} records[maxn], tmp;
bool cmp(const record &a, const record &b) {
if (a.name != b.name) return a.name < b.name;
else if (a.month != b.month) return a.month < b.month;
else if (a.day != b.day) return a.day < b.day;
else if (a.hour != b.hour) return a.hour < b.hour;
else if (a.minute != b.minute) return a.minute < b.minute;
}
int toll[24] = {0};
void get_ans(int on, int off, int &time, int &money) {
tmp = records[on];
while (tmp.day < records[off].day || tmp.hour < records[off].hour || tmp.minute < records[off].minute) {
++time;
money += toll[tmp.hour];
++tmp.minute;
if (tmp.minute == 60) {
tmp.minute = 0;
tmp.hour++;
}
if (tmp.hour == 24) {
tmp.hour = 0;
tmp.day++;
}
}
}
int main()
{
for (int i = 0; i < 24; ++i) {
scanf("%d", &toll[i]);
}
int n;
scanf("%d", &n);
for (int i = 0; i < n; ++i) {
std::cin >> records[i].name;
scanf("%d:%d:%d:%d", &records[i].month, &records[i].day, &records[i].hour, &records[i].minute);
std::cin >> records[i].word;
}
std::sort(records, records + n, cmp);
// 筛选有效的
int on = 0, off, next; // on和off是配对的一条,next为下一个用户
while (on < n) {
int needPrint = 0; // 记录是否找到有效记录 2为找到
next = on; // 从on开始扫描
while (next < n && records[next].name == records[on].name) {
if (needPrint == 0 && records[next].word == "on-line") {
needPrint = 1;
} else if (needPrint == 1 && records[next].word == "off-line") {
needPrint = 2;
}
++next;
}
if (needPrint < 2) {
on = next;
continue;
}
int allMoney = 0;
printf("%s %02d\n", records[on].name.c_str(), records[on].month);
while (on < next) {
while (on < next -1 && !(records[on].word == "on-line" && records[on + 1].word == "off-line")) {
++on;
}
off = on + 1;
if (off == next) {
on = next;
break;
}
printf("%02d:%02d:%02d ", records[on].day, records[on].hour, records[on].minute);
printf("%02d:%02d:%02d ", records[off].day, records[off].hour, records[off].minute);
int time = 0, money = 0;
get_ans(on, off, time, money);
allMoney += money;
printf("%d $%.2f\n", time, money / 100.0); // cents -> dollar
on = off + 1;
}
printf("Total amount: $%.2f\n", allMoney / 100.0);
}
return 0;
}
小结
- 先伪码大概思路模块化,然后再一个模块一个模块的去攻克。
1 dollar = 100 cents
PS:
稍微有点畏难,有点怕麻烦吧,还觉得自己的思路有点暴力,但是看了解答好像也是暴力的方式。然后以后要是没思路的话先看思路然后自己写写看,实在不行再看解答。