[C++知识库]牛客练习赛96 B

题目
题意: 求一颗n层的满k叉树，求任意两点之间距离和等于多少.
思路: 可以暴力也可以推公式，不会推。暴力做法为，对于同一层的每条边，贡献相同。而该边的贡献为cnt左边的点 * cnt右边的点.

时间复杂度: O(n)
代码:

// Problem: 小y的树
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/11186/B
// Memory Limit: 524288 MB
// Time Limit: 2000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<complex>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<unordered_map>
#include<list>
#include<set>
#include<queue>
#include<stack>
#define OldTomato ios::sync_with_stdio(false),cin.tie(nullptr),cout.tie(nullptr)
#define fir(i,a,b) for(int i=a;i<=b;++i) 
#define mem(a,x) memset(a,x,sizeof(a))
#define p_ priority_queue
// round() 四舍五入 ceil() 向上取整 floor() 向下取整
// lower_bound(a.begin(),a.end(),tmp,greater<ll>()) 第一个小于等于的
// #define int long long //QAQ
using namespace std;
typedef complex<double> CP;
typedef pair<int,int> PII;
typedef long long ll;
// typedef __int128 it;
const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const ll inf = 1e18;
const int N = 2e5+10;
const int M = 1e6+10;
const int mod = 1e9+7;
const double eps = 1e-6;
inline int lowbit(int x){ return x&(-x);}
template<typename T>void write(T x)
{
    if(x<0)
    {
        putchar('-');
        x=-x;
    }
    if(x>9)
    {
        write(x/10);
    }
    putchar(x%10+'0');
}
template<typename T> void read(T &x)
{
    x = 0;char ch = getchar();ll f = 1;
    while(!isdigit(ch)){if(ch == '-')f*=-1;ch=getchar();}
    while(isdigit(ch)){x = x*10+ch-48;ch=getchar();}x*=f;
}
#define int long long
int n,m,k,T;
int f[M];
int s[M];
void solve()
{
   read(n); read(k);
   int sum = 0;
   int ans = 0;
   f[0] = 1;
   s[0] = 1;
   for(int i=1;i<=n;++i) f[i] = f[i-1] * k % mod,s[i] = (s[i-1] + f[i]) % mod;
   sum = s[n-1];
   for(int i=1;i<=n-1;++i)
   {
   	  ans = (ans + f[n-i] * s[i-1] % mod * (sum-s[i-1]) % mod) % mod;
   }
   ans = (ans + mod) % mod;
   write(ans);
}
signed main(void)
{  
   T = 1;
   // OldTomato; cin>>T;
   // read(T);
   while(T--)
   {
   	 solve();
   }
   return 0;
}