Gym - 100637A Nano alarm-clocks 模拟

题意：有n个时钟，只能顺时针拨，问使所有时间相同的最小代价是多少

思路：将时间排序，枚举拨动到每一个点的时间就好了，容易证明最终时间一定是其中之一

 #include <iostream>
 #include <cstdio>
 #include <fstream>
 #include <algorithm>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <string>
 #include <cstring>
 #include <map>
 #include <stack>
 #include <set>
 #define LL long long
 #define eps 1e-8
 #define INF 0x3f3f3f3f
 #define MAXN 200005
 #define U 1000000
 using namespace std;
 LL t[MAXN], sum1[MAXN], sum2[MAXN];
 int main()
 {
 #ifndef ONLINE_JUDGE
     freopen("in.txt", "r", stdin);
     //freopen("out.txt", "w", stdout);
 #endif // OPEN_FILE
     int n;
     memset(t, , sizeof(t));
     memset(sum1, , sizeof(sum1));
     memset(sum2, , sizeof(sum2));
     LL x, y, z;
     scanf("%d", &n);
     for(int i = ; i <= n; i++){
         scanf("%I64d%I64d%I64d", &x, &y, &z);
         t[i] = (x * U + y) * U + z;
     }
     sort(t + , t + n + );
     for(int i = ; i <= n; i++){
         sum1[i] = sum1[i - ] + t[i];
     }
     for(int i = n; i >= ; i--){
         sum2[i] = sum2[i + ] + t[i];
     }
     LL ans;
     bool flag = false;
     LL ALL = 1000000LL * 1000000LL * 12LL;
     for(int i = ; i <= n; i++){
         LL res = (i - ) * t[i] - sum1[i - ] + ALL * (n - i) - (sum2[i + ] - (n - i) * t[i]);
         if(!flag){
             ans = res;
             flag = true;
         }
         else{
             ans = min(ans, res);
         }
     }
     LL s = ans % U;
     ans = ans / U;
     LL m = ans % U;
     ans = ans / U;
     LL h = ans;
     printf("%I64d %I64d %I64d\n", h, m, s);
 }