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

时间：2022-07-12 21:59:06

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，问你总计转多少时间，可以使得所有表的时间一样

注意，只能往前拨

题解：

先排序，然后维护前缀和，对于每一个在他前面的表，可能时间只能转到和他的时间一样的时候

在他后面的表，就转到他的时间+12h就好了

然后跑一遍

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 long long ll;

using namespace std;

//freopen("D.in","r",stdin);

//freopen("D.out","w",stdout);

#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)

#define test freopen("test.txt","r",stdin)

const int maxn=;

#define mod 1000000007

#define eps 1e-9

const int inf=0x3f3f3f3f;

const ll infll = 0x3f3f3f3f3f3f3f3fLL;

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;

}

//**************************************************************************************

unsigned long long ti[maxn];

unsigned long long sum[maxn];

int main()

{

    int n=read();

    for(int i=;i<=n;i++)

    {

        unsigned long long x=read(),y=read(),z=read();

        ti[i]=z+y*1000000LL+x*1000000LL*1000000LL;

    }

    sort(ti+,ti++n);

    for(int i=;i<=n;i++)

        sum[i]=sum[i-]+ti[i];

    unsigned long long ans=;

    for(int i=;i<=n;i++)

    {

        unsigned long long num=i*ti[i]-sum[i];

        num+=(12LL*1000000LL*1000000LL+ti[i])*(n-i)-sum[n]+sum[i];

        if(i==)

            ans=num;

        else

            ans=min(ans,num);

    }

    unsigned long long hour=1000000LL*1000000LL;

    unsigned long long h=ans/(1000000LL*1000000LL);

    unsigned long long m=(ans-h*(1000000LL*1000000LL))/1000000LL;

    unsigned long long s=ans-h*1000000LL*1000000LL-m*1000000LL;

    cout<<h<<" "<<m<<" "<<s<<endl;

}

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