Build Tree

题目描述：

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∑i=0Nd(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.

输入：

The input file contains 2 lines.

The first line contains 4 integers,these respectively is k,m,n,p。(2 ≤ k ≤200000,2 ≤ p≤ 1015)

The second line contains k integers,represent the set S，the elements in the set guarantee less than or equal to 1015.

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

输出：

The output file contains an integer.represent the answer.

样例输入：

5 2 3 10
1 2 3 4 5

样例输出：

6

题目大意是，给出n叉树，并给出层数，要求在以下的数字集合中选出数字，组成满n叉树，使得每一个结点到根节点的距离之和最大
根据贪心策略，使总距离最大，那么应该将较小的边放在尽可能靠前的位置
代码如下：

#include<stdio.h>
#include<string.h>
#include<algorithm>
long long a[300000]; 
long long s[300000];
int main()
{
	long long k,n,m,p;
	while(~scanf("%lld%lld%lld%lld",&k,&m,&n,&p))
	{
		int i,j;
		memset(a,0,sizeof(a));
		memset(s,0,sizeof(s));
		for(i=0;i<k;i++)
		{
			scanf("%lld",&a[i]);
		}
		std::sort(a,a+k);
		int ceng=n;
		int count=0;
		for(i=2;i<=m;i++)
		{
			s[i]=(s[i-1]*n)%p;
			for(j=0;j<ceng;j++)
			{
				s[i]=(s[i]+a[count++])%p;
			}
			ceng=ceng*n;
		}
		long long sum=0;
		for(i=2;i<=m;i++)
		{
			sum=(sum+s[i])%p;
		}
		printf("%lld\n",sum);
	}
	return 0;
}