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
1 #include <iostream> 2 #include <cstdio> 3 #include <fstream> 4 #include <algorithm> 5 #include <cmath> 6 #include <deque> 7 #include <vector> 8 #include <queue> 9 #include <string> 10 #include <cstring> 11 #include <map> 12 #include <stack> 13 #include <set> 14 #define LL long long 15 #define eps 1e-8 16 #define INF 0x3f3f3f3f 17 #define MAXN 200005 18 #define U 1000000 19 using namespace std; 20 LL t[MAXN], sum1[MAXN], sum2[MAXN]; 21 int main() 22 { 23 #ifndef ONLINE_JUDGE 24 freopen("in.txt", "r", stdin); 25 //freopen("out.txt", "w", stdout); 26 #endif // OPEN_FILE 27 int n; 28 memset(t, 0, sizeof(t)); 29 memset(sum1, 0, sizeof(sum1)); 30 memset(sum2, 0, sizeof(sum2)); 31 LL x, y, z; 32 scanf("%d", &n); 33 for(int i = 1; i <= n; i++){ 34 scanf("%I64d%I64d%I64d", &x, &y, &z); 35 t[i] = (x * U + y) * U + z; 36 } 37 sort(t + 1, t + n + 1); 38 for(int i = 1; i <= n; i++){ 39 sum1[i] = sum1[i - 1] + t[i]; 40 } 41 for(int i = n; i >= 1; i--){ 42 sum2[i] = sum2[i + 1] + t[i]; 43 } 44 LL ans; 45 bool flag = false; 46 LL ALL = 1000000LL * 1000000LL * 12LL; 47 for(int i = 1; i <= n; i++){ 48 LL res = (i - 1) * t[i] - sum1[i - 1] + ALL * (n - i) - (sum2[i + 1] - (n - i) * t[i]); 49 if(!flag){ 50 ans = res; 51 flag = true; 52 } 53 else{ 54 ans = min(ans, res); 55 } 56 } 57 LL s = ans % U; 58 ans = ans / U; 59 LL m = ans % U; 60 ans = ans / U; 61 LL h = ans; 62 printf("%I64d %I64d %I64d ", h, m, s); 63 }