链接：https://www.nowcoder.com/acm/contest/147/E
来源：牛客网
时间限制：C/C++ 1秒，其他语言2秒
空间限制：C/C++ 262144K，其他语言524288K
64bit IO Format: %lld
题目描述
Niuniu likes to play OSU!
We simplify the game OSU to the following problem.
Given n and m, there are n clicks. Each click may success or fail.
For a continuous success sequence with length X, the player can score X^m.
The probability that the i-th click success is p[i]/100.
We want to know the expectation of score.
As the result might be very large (and not integral), you only need to output the result mod 1000000007.
输入描述:
The first line contains two integers, which are n and m. The second line contains n integers. The i-th integer is p[i]. 1 <= n <= 1000 1 <= m <= 1000 0 <= p[i] <= 100
输出描述:
You should output an integer, which is the answer.
示例1
输入
3 4 50 50 50
输出
750000020
说明
000 0 001 1 010 1 011 16 100 1 101 2 110 16 111 81 The exact answer is (0 + 1 + 1 + 16 + 1 + 2 + 16 + 81) / 8 = 59/4. As 750000020 * 4 mod 1000000007 = 59 You should output 750000020.
备注:
If you don't know how to output a fraction mod 1000000007, You may have a look at https://en.wikipedia.org/wiki/Modular_multiplicative_inverse
题意：有一个长度为n只包含01的序列，每个位置为1的可能性给出，每段连续的长度为x的序列可以得到分数x^m，问最后得到总分期望。
思路：
BZOJ 3450 平方的版本。
BZOJ 4318 立方的版。
hdu 4625 JZPTREE，组合数维护起来比较简单，最后转回次方即可。
对于m等于2的情况 (x+1)^2 = x^2+2x+1 .每次只需要在前面平方期望的基础上加上（前面长度期望的两倍加上1）*现在这个点为1的概率。
m=3同理。
但是m<=1000不好维护
在hdu4625中给出了把幂转换为组合数来维护的方法,再通过第二类Stirling数变回来。C(n,k) = C(n-1,k) + C(n-1,k-1)比较好转移。
hdu4625代码
#include<iostream>
#include<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn =5e4+10;
const int maxk = 505;
const int mod = 10007;
int n,k;
vector<int>g[maxn];
int dp[maxn][maxk];
int fac[maxk];
int s[maxk][maxk];
void init()
{
fac[0] = 1;
for(int i = 1;i<maxk;i++)
fac[i] = (fac[i-1] * i)%mod;
memset(s,0,sizeof(s));
s[1][1]=1;
for(int i=2;i<maxk;i++)
for(int j=1;j<=i;j++)
{
s[i][j]=(s[i-1][j-1]+(j*s[i-1][j])%mod)%mod;
}
}
void dfs(int u,int f)
{
int v;
for(int i = 0;i<g[u].size();i++)
{
v = g[u][i];
if(v==f)continue;
dfs(v,u);
dp[u][0] = (dp[u][0] + dp[v][0])%mod;
for(int j = 1;j<=k;j++)
{
dp[u][j] = ((dp[u][j] + dp[v][j])%mod + dp[v][j-1])%mod;
}
}
dp[u][0]++;
dp[u][0] %= mod;
}
void dfs2(int u,int f)
{
int v;
for(int i = 0;i<g[u].size();i++)
{
v = g[u][i];
if(v==f)continue;
for(int j = k;j>0;j--)
{
int p = ((dp[u][j] - dp[v][j] + mod)%mod - dp[v][j-1] +mod)%mod;
int q = ((dp[u][j-1] - dp[v][j-1] + mod)%mod - (j-2>=0?dp[v][j-2]:0) + mod)%mod;
dp[v][j] = (dp[v][j] + p + q)%mod;
}
dp[v][0] = (dp[v][0] + (dp[u][0] - dp[v][0] + mod)%mod)%mod;
dfs2(v,u);
}
}
int main()
{
init();
int t;
scanf("%d",&t);
while(t--)
{
memset(dp,0,sizeof(dp));
scanf("%d%d",&n,&k);
for(int i = 0;i<=n;i++)
{
g[i].clear();
}
int u,v;
for(int i=0;i<n-1;i++)
{
scanf("%d%d",&u,&v);
g[u].push_back(v);
g[v].push_back(u);
}
dfs(1,-1);
dfs2(1,-1);
int ans;
for(int u = 1;u<=n;u++)
{
ans = 0;
for(int i = 0;i<=k;i++)
{
ans = (ans + ((s[k][i] * fac[i])%mod * dp[u][i])%mod)%mod;
}
printf("%d\n",ans);
}
}
}这一题中也可以这样做，标程代码中组合数的维护方法学习了。
#include <bits/stdc++.h>
using namespace std;
int n, m, mod = 1000000007;
long long p[1020];
long long a[1020];
long long b[1020];
long long s[1020][1020];
long long f[1020];
int main() {
scanf("%d%d", &n, &m);
s[0][0] = 1;
for (int i = 0; i <= m; i++) {
for (int j = 1; j <= i; j++) {
s[i][j] = (s[i - 1][j - 1] + j * s[i - 1][j]) % mod;
}
}
for (int i = f[0] = 1; i <= m; i++) {
f[i] = f[i - 1] * i % mod;
}
//维护组合数，a[i]为以i结尾的一段连续combo的组合数期望，b[i]为前面总的组合数期望
for (int i = 1; i <= n; i++) {
scanf("%lld", &p[i]);
p[i] = p[i] * 570000004 % mod;
for (int j = m; j > 0; j--) {
a[j] += a[j - 1];
a[j] *= p[i];
a[j] %= mod;
b[j] += a[j];
b[j] %= mod;
}
a[0] = p[i];
b[0] += a[0];
b[0] %= mod;
}
long long z = 0;
for (int i = 1; i <= m; i++) {
z = (z + b[i - 1] * f[i] % mod * s[m][i]) % mod;
}
printf("%lld\n", z);
return 0;
}这一题还可以用更简单的方法去做，枚举每一个区间为完整的一个combo得到的值乘以概率全部加到ans上即可。