Nyoj 476 who is the hero (unique factor decomposition theorem)

Title Description

http://acm.nyist.net/JudgeOnline/problem.php?pid=476

10 Mathematicians (No. 0-9) flew over the Pacific Ocean by balloon. When crossing the equator, they decided to celebrate the feat. So they opened a bottle of champagne. Unfortunate

, the Cork has a hole in the balloon, the hydrogen leaks, the balloon starts to fall, and it's going to fall into the sea, and everyone is going to be eaten by sharks.

But there are chance--if one of them sacrifices himself to jump, his friends will live a little longer. But there is still a problem--who

Jump down? So they think of a very fair way to solve this problem--first, each person writes an integer AI;

Figure out the number of a1xa2xa3xa4x......xa10 of the product of N. For example, 6 of the approximate 4 (1, 2, 3, 6), then n is 4. The Sacrifice from

The hero of his own will be determined by the number of bits in N (the number of single digits of n is to jump). Your task is to find out N.

Input

T (t<=10) group test data.
10 integer ai (1≤ai≤10000).

Output

Single digit of n

Sample input

11 2 6 1 3 1 1 1 1 1

Sample output

9

Topic Analysis:

The only factor decomposition problem, for any number n, can be expressed as: n=p[0]^a[0]*... *p[i]^a[i]*......*p[n]^a[n], where P[i] is a prime, a[i] is an integer > 0, while the sum of the approximate number of n is, sum= (a[0]+1) *.....* (a[i]+1)* ... (a[n]+1).

AC Code:

/** * @xiaoran * unique prime factor decomposition, the approximate number of n is pi (a[i]+1) *res= (a[0]+1) * (a[1]+1) *...* (a[n]+1), where a[i] are prime numbers */#include <iostream> #include <cstdio> #include <map> #include <cstring> #include <string> #include <algorithm > #include <queue> #include <vector> #include <stack> #include <cstdlib> #include <cctype    > #include <cmath> #define LL long longusing namespace Std;int a[10000];int main () {int t,n;    cin>>t;        while (t--) {int res=1,x,k,m=0;        memset (A,0,sizeof (a));            for (int i=0;i<10;i++) {cin>>n;            if (m<n) m=n;//record maximum value x=n;                    for (int i=2;i<=n;i++) {while (x%i==0) {//record the number of times each element factor appears a[i]++;                X/=i;            } if (x==1) break;        }} for (int i=2;i<=m;i++) {res*= (a[i]+1);    } cout<<res%10<<endl; }return 0;}

Nyoj 476 who is the hero (unique factor decomposition theorem)