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\n",(ans/t)/t,(ans/t)%t,ans%t);

60     return 0;

61 }

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