Codeforces Gym 100637A A. Nano alarm-clocks 前缀和处理

A. Nano alarm-clocks

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/gym/100637/problem/A

Description

An old watchmaker has n stopped nano alarm-clocks numbered with integers from 1 to n. Nano alarm-clocks count time in hours, and in one hour there are million minutes, each minute lasting a million seconds. In order to repair them all the watchmaker should synchronize the time on all nano alarm-clocks. In order to do this he moves clock hands a certain time forward (may be zero time). Let’s name this time shift a transfer time.

Your task is to calculate the minimal total transfer time required for all nano alarm-clocks to show the same time.

Input

The first line contains a single integer n — the number of nano alarm-clocks (2 ≤ n ≤ 105). In each i-th of the next n lines the time h, m,s, shown on the i-th clock. Integers h, m and s show the number of hours, minutes and seconds respectively. (0 ≤ h < 12, 0 ≤ m < 106,0 ≤ s < 106).

Output

Output three integers separated with spaces h, m and s — total minimal transfer time, where h, m and s — number of hours, minutes and seconds respectively (0 ≤ m < 106, 0 ≤ s < 106).

Sample Input

2
10 0 0
3 0 0

Sample Output

5 0 0

HINT

题意

给你n个时钟，只可向前拨 问你总计拨多少时间，可以使得所有表的时间一样

题解：

排序+维护前缀和，暴力出最小的就OK

代码

 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <ctime>
 #include <iostream>
 #include <algorithm>
 #include <set>
 #include <vector>
 #include <sstream>
 #include <queue>
 #include <typeinfo>
 #include <fstream>
 #include <map>
 #include <stack>
 typedef __int64 ll;
 using namespace std;
 inline ll read()
 {
     ll x=,f=;
     char ch=getchar();
     while(ch<''||ch>'')
     {
         if(ch=='-')f=-;
         ch=getchar();
     }
     while(ch>=''&&ch<='')
     {
         x=x*+ch-'';
         ch=getchar();
     }
     return x*f;
 }
 //**************************************************************************************
 ll t=;
 int n;
 ll sum[];
 ll a[];
 int main()
 {

     scanf("%d",&n);
     for(int i=; i<=n; i++)
     {
         ll h,m,s;
         cin>>h>>m>>s;
         a[i]=s+m*t+t*t*h;
     }
     sort(a+,a+n+);
     for(int i=;i<=n;i++)
         sum[i]=sum[i-]+a[i];
         ll tt=*t*t;
         ll ans=tt*;//此处无穷大就好了

     for(int i=n;i>=;i--)
     {
         ll xx=(a[i]*(i-)-sum[i-]+(a[i]+tt)*(n-i)-(sum[n]-sum[i]));
         ans=min(xx,ans);
     }
     printf("%I64d %I64d %I64d\n",(ans/t)/t,(ans/t)%t,ans%t);
     return ;
 }