Prefix and suffix Product

Description:

Give the series A1, A2,..., An, and set

Bi = (A1 * A2 * A3... An)/AI Mod (109 + 7)

Now we need to calculate all bi

Input:

The input contains multiple groups of test data. For each group of data, the length of the series is represented in 1st rows and 1 integer N (1 ≤ n ≤ 100,000. 2nd

Row, N integers A1, A2,..., an (1 ≤ AI ≤ 109), indicating the given series. The input end with a value of 0.

Output:

For each group of data, a row is output. N integers are separated by spaces, indicating the calculated b1, b2,..., bn.

Sample input:

3

1 2 3

0

Sample output:

6

3

2

 

Difficult to understand

The data given by the question is very big. If we continue to multiply, it will certainly overflow. So here we use a remainder allocation law (a * B) % C = (a % C * B % C) % C

However, this only produces A1 * a2...... an. Because there is no allocation rate in Division, prefix and suffix product are used to solve this problem.

That is, if n = 2, Bi = (A1 * A2 * A3 * A4 * A5) % AI => pre [I] = a1 % 1000000007 suff [I] = (A3 * A4 * A5) % 1000000007 Bi = pre [I] * suff [I]

The Code is as follows:

 

 1 #include<stdio.h> 2 #include<string.h> 3 #include<stdlib.h> 4 using namespace std; 5 long long a[100010],pre[100010],suff[100010]; 6 const int mod=1000000007; 7 int main() 8 { 9     int n,i;10     while(scanf("%d",&n)&&n)11     {12         memset(a,0,sizeof(a));13         for(i=1;i<=n;i++)14             scanf("%lld",&a[i]);15         pre[0]=1;16         for(i=1;i<=n;i++)17             pre[i]=(pre[i-1]*a[i])%mod;18         suff[n+1]=1;19         for(i=n;i>=1;i--)20             suff[i]=(suff[i+1]*a[i])%mod;21         for(i=1;i<=n;i++)22             printf("%lld%c",(pre[i-1]*suff[i+1])%mod,i==n?‘\n‘:‘ ‘);23     }24     return 0;25 }