洛谷P3509 Frog

题目
首先分析数据范围发现m很大，所以线性做法肯定不行，因此考虑倍增，即预处理出每个点跳1次后的位置。然后只用两个数组类似于快速幂，推出每个点跳m次后的位置。

预处理离每个点第k小的点，可以用长度为k的尺子尺取。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define int long long
#define N 1001011
using namespace std;
int n, k, m;
int data[N], jum[N], ans[N], temp[N];
inline void init()				
{								
	scanf("%lld%lld%lld", &n, &k, &m);
	for (int i = 1; i <= n; i++)
		scanf("%lld", &data[i]), ans[i] = i;
	jum[1] = k + 1;
	int l = 1, r = k + 1; 
	for (int i = 2; i <= n; i++)
	{
		while (r + 1 <= n && (data[r + 1] - data[i]) < (data[i] - data[l]) ) l++, r++;
		if (data[r] - data[i] > data[i] - data[l])
			jum[i] = r;
		else
			jum[i] = l;
	}
}
signed main()
{            
	init();  
	while (m)
	{        
		if (m & 1)
			for (int i = 1; i <= n; i++)
				ans[i] = jum[ans[i]];
		m >>= 1;
		for (int i = 1; i <= n; i++)
			temp[i] = jum[i];
		for (int i = 1; i <= n; i++)
			jum[i] = temp[temp[i]];
	}
	for (int i = 1; i <= n; i++)
		printf("%d ", ans[i]);
	return 0;
}