牛客多校赛第九场 E Music Game 概率计算
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有n次点击每次点击成功的概率是p[i]/100，连续点击成功X次会获得X^m的分数，求分数的期望。
题解来源
作者：Cload9
链接：https://www.nowcoder.com/discuss/94910?type=101&order=0&pos=1&page=0
来源：牛客网
首先，预处理出每一段一直成功的概率。
然后我们可以枚举i,j表示在[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)。
#include
#define ll long long
#define mod 1000000007
using namespace std;
ll _pow(ll a,ll b)
{
ll res=1;
while(b>0)
{
if(b&1)
res=res*a%mod;
a=a*a%mod;
b>>=1;
}
return res;
}
ll p[1010][1010];
ll q[1010];
ll a[1010];
int main()
{
ll inv=_pow(100,1000000005);
ll n,m;
scanf("%lld%lld",&n,&m);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]),q[i]=_pow(i,m);
for(int i=1;i<=n;i++)
{
p[i][i]=a[i]*inv%mod;
for(int j=i+1;j<=n;j++)
p[i][j]=p[i][j-1]*a[j]%mod*inv%mod;
}
ll ans=0;
for(int i=0;i