2019CCPC湖南全国邀请赛（广东省赛、江苏省赛）（B - Build Tree）（找规律）

2019CCPC湖南全国邀请赛（广东省赛、江苏省赛）（B - Build Tree）（找规律）

Time limit1500 ms
Memory limit65536 kB

Description

You need to construct a full $n$-ary tree(n叉树) with $m$ layers.All the edges in this tree have a weight.But this weight cannot be chosen arbitrarily you can only choose from set $S$，the size of $S$ is $k$，each element in the set can only be used once.Node 0 is the root of tree.

We use $d(i)$ for the distance from root to node $i$.Our goal is to minimize the following expression:
$$min\sum _{i=0}^{N}d(i)$$
Please find the minimum value of this expression and output it.Because it may be a big number,you should output the answer modul $p$.

Input

The input file contains 2 lines.

The first line contains 4 integers,these respectively is $k,m,n,p。(2\le k\le 200000,2\le p\le 10^{15})$

The second line contains $k$ integers,represent the set S，the elements in the set guarantee less than or equal to $10^{15}$.

We guarantee that $k$ is greater than or equal to the number of edges.

Output

The output file contains an integer.represent the answer.

Sample Input

5 2 3 10
1 2 3 4 5

Sample Output

6

题意

给你一个序列，长度为k，让你利用这些值当作边的权值建立一颗m层的满n叉树，每个点到根节点的距离是路径的权值之和，然后问你每一个边到根节点的距离之和。

本来就是一道规 （S） 律 （B） 题，可是明明思路很清楚，却总不知道怎么敲下那代码，于是就有了这篇整理。

题解

一切尽在图中

代码

#include <bits/stdc++.h>
#define maxn 200005
#define _for(i, a) for(int i = 0; i < (a); ++i)
using namespace std;
typedef long long LL;

LL n, m, k, p;
LL a[maxn];
LL ans;
LL num[maxn];

void init() {
    
    
	ans = 0;
	num[0] = 0;
	num[1] = 0;
	for (LL i = 2, t = n; i <= m; ++i, t *= n) num[i] = num[i - 1] + t, num[i] %= p;
}

void sol() {
    
    
	init();
	_for(i, k) scanf("%lld", &a[i]);
	sort(a, a + k);
	for (int i = 1; i < k; ++i) a[i] += a[i - 1], a[i] %= p;
	if (m == 1) ans = 0;
	else {
    
    
		for (LL i = 2; i <= m; ++i) {
    
    
			LL _sum = a[num[i] - 1] + p - (i == 2 ? 0 : a[num[i - 1] - 1]); _sum %= p;
			LL times = (num[m - i + 1] + 1) % p;
			ans += _sum * times, ans %= p;
		}
	}
	cout << ans << "\n";
}

int main() {
    
    
	while (cin >> k >> m >> n >> p) {
    
    
		sol();
	}
	return 0;
}