链接：https://www.nowcoder.com/acm/contest/147/E
来源：牛客网
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
#include<bits/stdc++.h>
#define ll long long
#define inf 0x3f3f3f3f
using namespace std;
const int maxn=1010;
const ll mo=1e9+7;
ll n,m,k;
ll a[maxn],dp[maxn][maxn],sum[maxn][maxn];
ll c[maxn];
ll ans,ct,cnt,tmp,flag;
ll power(ll a,ll n) //a的n次方mod
{
ll ans=1;
a=a%mo;
while (n)
{
if(n&1) ans=(ans*a)%mo;
n>>=1;
a=(a*a)%mo;
}
return ans;
}
int main()
{
int T,cas=1;
while(scanf("%lld %lld",&n,&m)!=EOF)
{
c[1]=1;
for(int i=2;i<=1000;i++) c[i]=power(i,m);
ans=0;
tmp=power(100,mo-2);
for(int i=1;i<=n;i++){
scanf("%lld",&a[i]);
}
a[0]=a[n+1]=0;
for(int i=1;i<=n;i++)
{
dp[i][i-1]=1;
for(int j=i;j<=n;j++)
{
dp[i][j]=dp[i][j-1]*a[j]%mo*tmp%mo;
ans=(ans+dp[i][j]*(100-a[i-1])%mo*tmp%mo*(100-a[j+1])%mo*tmp%mo*c[j-i+1]%mo)%mo;
}
}
ans=(ans+mo)%mo;
printf("%lld\n",ans);
// if(flag) puts("Yes"); else puts("No");
}
return 0;
}