E - GukiZ and GukiZiana 思路:分块, 块内二分 代码: #pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pi acos(-1.0) #define LL long long //#define mp m…
E. GukiZ and GukiZiana Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/551/problem/E Description Professor GukiZ was playing with arrays again and accidentally discovered new function, which he called GukiZiana. For given a…
GukiZ and GukiZiana Time Limit: 10000ms Memory Limit: 262144KB This problem will be judged on CodeForces. Original ID: 551E64-bit integer IO format: %I64d Java class name: (Any) Professor GukiZ was playing with arrays again and accidentally di…
Problem E. GukiZ and GukiZiana Solution: 先分成N=sqrt(n)块,然后对这N块进行排序. 利用二分查找确定最前面和最后面的位置. #include <bits/stdc++.h> using namespace std; typedef long long ll; vector<]; ll add[], a[], pos[]; ll n, q, bk, N; bool cmp( int x, int y ) { if( a[x] == a[y]…
E. GukiZ and GukiZiana time limit per test 10 seconds memory limit per test 256 megabytes input standard input output standard output Professor GukiZ was playing with arrays again and accidentally discovered new function, which he called GukiZiana. F…
