2019CCPC湖南全國邀請賽（廣東省賽、江蘇省賽）（B - Build Tree）（找規律）

2019CCPC湖南全國邀請賽（廣東省賽、江蘇省賽）（B - Build Tree）（找規律）

Time limit1500 ms
Memory limit65536 kB

Description

You need to construct a full $n$-ary tree(n叉樹) 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:
$min∑_{i=0}^N{d(i)}$
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 $k,m,n,p。(2 ≤ k ≤200000,2 ≤ p≤ 10^{15})$

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

題意

給你一個序列，長度爲k，讓你利用這些值當作邊的權值建立一顆m層的滿n叉樹，每個點到根節點的距離是路徑的權值之和，然後問你每一個邊到根節點的距離之和。

本來就是一道規 （S） 律 （B） 題，可是明明思路很清楚，卻總不知道怎麼敲下那代碼，於是就有了這篇整理。

題解

一切盡在圖中

代碼

#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;
}