Music Game

链接：https://www.nowcoder.com/acm/contest/147/E
来源：牛客网

Music Game

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

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 81The 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

分析：第i次点击时的长度，设第i-1次点击时的长度x

len=(1-pi)*0+pi(x+1)

第i次点击时的得分，dp[i]=(1-pi)*0+pi*(x+1)^m

=pi*(C(m,0)*x^m+C(m,1)*x^(m-1)+···+C(m,m-1)*x^(m-(m-1))+C(m,m)*x^(m-m))

=pi*(dp[i-1]+C(m,1)*x^(m-1)+···+C(m,m-1)*x^(m-(m-1))+C(m,m)*x^(m-m)

所以需要维护x^(m-1)···x好难，不懂

所以暴力YY了

#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;//枚举哪一段是1for(int i=1;i<=n;i++){dp[i][i-1]=1;//没选段的时候期望是1for(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;//[i,j]是1必须保证前一个后一个都是0，不然就不是[i,j]是1，就比[i,j]大了}}ans=(ans+mo)%mo;printf("%lld\n",ans);//  if(flag) puts("Yes"); else puts("No");}return 0;
}

Music Game

题目描述

输入描述:

输出描述:

输入

输出

说明

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