牛客多校第九场 E Music Game 概率计数
链接：https://www.nowcoder.com/acm/contest/147/E
来源：牛客网
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
复制
3 4 50 50 50
输出
复制
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
本题题意：
给出长度为n的序列a，表示每个点成功的概率，
从1到n的分数计算方法是，若成功则获得连续成功的长度的m次方的分数，求最后的分数期望。
首先，预处理出每一段一直成功的的概率。
然后，枚举i,j表示出在[i+1,j-1]区间内成功连击，并且由题目可知要计算上这段分数必须在i,j失败，结束连击
这段的期望就是（i失败的概率）*（j失败的概率）*（[i+1,j-1]都成功的概率）
为了方便计算，把0,n+1两个点的成功概率都设为0
#include
using namespace std;
const long long mod=1000000007,inv=570000004;
long long i,i0,n,m,T,a[1005],pre[1005][1005],mpow[1005],ans;
long long qpow(long long a,long long b,long long mod)
{long long r=1,t=a; while(b){if(b&1)r=(r*t)%mod;b>>=1;t=(t*t)%mod;}return r;}
int main()
{
scanf("%lld %lld",&n,&m);
a[0]=a[n+1]=0,ans=0;
for(i=1;i<=n;i++)scanf("%lld",&a[i]),mpow[i]=qpow(i,m,mod);
for(i=1;i<=n;i++)
{
pre[i][i]=a[i]*inv%mod;
for(i0=i+1;i0<=n;i0++)pre[i][i0]=pre[i][i0-1]*a[i0]%mod*inv%mod;
}
for(i=0;i<=n;i++)
{
for(i0=i+2;i0<=n+1;i0++)
{
ans+=pre[i+1][i0-1]*mpow[i0-i-1]%mod*(100-a[i])%mod*inv%mod*(100-a[i0])%mod*inv%mod;
ans%=mod;
}
}
printf("%lld\n",ans);
return 0;
}