CF Gym 100637A Nano alarm-clocks

题意：给你一些钟的时间，只可以往后调， 问最少调的时间总和是多少

题解：因为肯定是调到某个出现过时间的，只要枚举时间，在维护一个前缀和快速计算出时间总和就行了。

#include<cstdio>
#include<cmath>
#include<vector>
#include<map>
#include<set>
#include<algorithm>
#define first fi
#define second se
using namespace std;
typedef long long ll;

const int maxn = 100005;
const int hmod = 12;
const ll mmod = 1e6;
const ll Fac = 1e12;
const ll tmod = 12e12;//ll

ll sum[maxn];
ll times[maxn];
int main()
{
    int n;
    scanf("%d",&n);

    for(int i = 1; i <= n; i++){
        int th,tm,ts;
        scanf("%d%d%d",&th,&tm,&ts);
        times[i] = th*Fac+tm*mmod+ts;
    }
    sort(times+1,times+1+n);
    for(int i = 1; i <= n; i++)
        sum[i] = sum[i-1]+times[i];
    ll ans = 0x7fffffffffffffffll;
    for(int i = 1; i <= n; i++){
        ll tmp = i*times[i] - sum[i];   //前面的钟拨到time(i)
        tmp += (tmod+times[i])*(n-i)-sum[n]+sum[i];//后面的钟拨到time(i)
        ans = min(ans,tmp);
    }

    int ansH,ansM,ansS;
    ansH = ans/Fac; ans -= ansH*Fac;
    ansM = (ans/mmod); ans -= ansM*mmod;
    ansS = ans;
    printf("%d %d %d",ansH,ansM,ansS);
    return 0;
}