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

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <ctime>
 5 #include <iostream>
 6 #include <algorithm>
 7 #include <set>
 8 #include <vector>
 9 #include <sstream>
10 #include <queue>
11 #include <typeinfo>
12 #include <fstream>
13 #include <map>
14 #include <stack>
15 typedef __int64 ll;
16 using namespace std;
17 inline ll read()
18 {
19     ll x=0,f=1;
20     char ch=getchar();
21     while(ch<'0'||ch>'9')
22     {
23         if(ch=='-')f=-1;
24         ch=getchar();
25     }
26     while(ch>='0'&&ch<='9')
27     {
28         x=x*10+ch-'0';
29         ch=getchar();
30     }
31     return x*f;
32 }
33 //**************************************************************************************
34 ll t=1000000;
35 int n;
36 ll sum[200000];
37 ll a[200000];
38 int main()
39 {
40 
41     scanf("%d",&n);
42     for(int i=1; i<=n; i++)
43     {
44         ll h,m,s;
45         cin>>h>>m>>s;
46         a[i]=s+m*t+t*t*h;
47     }
48     sort(a+1,a+n+1);
49     for(int i=1;i<=n;i++)
50         sum[i]=sum[i-1]+a[i];
51         ll tt=12*t*t;
52         ll ans=tt*1000000;//此处无穷大就好了
53         
54     for(int i=n;i>=1;i--)
55     {
56         ll xx=(a[i]*(i-1)-sum[i-1]+(a[i]+tt)*(n-i)-(sum[n]-sum[i]));
57         ans=min(xx,ans);
58     }
59     printf("%I64d %I64d %I64d
",(ans/t)/t,(ans/t)%t,ans%t);
60     return 0;
61 }

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