The Factor(數論規律題+求質因子)



Link:http://acm.hdu.edu.cn/showproblem.php?pid=5428



The Factor

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 688    Accepted Submission(s): 229


Problem Description
There is a sequence of  n  positive integers. Fancycoder is addicted to learn their product, but this product may be extremely huge! However, it is lucky that FancyCoder only needs to find out one factor of this huge product: the smallest factor that contains more than 2 factors(including itself; i.e. 4 has 3 factors so that it is a qualified factor). You need to find it out and print it. As we know, there may be none of such factors; in this occasion, please print -1 instead. 
 

Input
The first line contains one integer  T (1T15) , which represents the number of testcases. 

For each testcase, there are two lines:

1. The first line contains one integer denoting the value of  n (1n100) .

2. The second line contains  n  integers  a1,,an (1a1,,an2×109) , which denote these  n  positive integers. 
 

Output
Print  T  answers in  T  lines.
 

Sample Input
  
  
  
2
3
1 2 3
5
6 6 6 6 6
 

Sample Output
  
  
  
6
4
 

Source
 

官方題解:

The Factor

對於每一個數字,它有用的部分其實只有它的所有質因子(包括相等的)。求出所有數的所有質因子中最小的兩個,相乘就是答案。如果所有數字的質因子個數不到兩個,那么就是無解。時間復雜度O(n*sqrt(a))O(nsqrt(a))

AC code:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <algorithm>
#define PI acos(-1.0)
#define LINF 1000000000000000000LL
#define eps 1e-8
#define LL long long
#define MAXN 1000010
#define MOD 9901
using namespace std;
const int INF=0x3f3f3f3f;
//******************************************
//素數篩選和合數分解
int cnt;//質因子個數
long long ans;
LL b[MAXN];
LL a[111];
int main()
{
int A,B,n;
int t,i;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
cnt=0;
for(i=1;i<=n;i++)
{
scanf("%I64d",&a[i]);
for(int j=2;(LL)j*j<=a[i];j++){
while(a[i]%j==0){
b[++cnt]=j;
a[i]/=j;
}
}
if(a[i]!=1){
b[++cnt]=a[i];
}
}
if(cnt>=2)
{
sort(b+1,b+cnt+1);
printf("%I64d\n",b[1]*b[2]);
}
else
{
printf("-1\n");
}
}
return 0;
}




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