light_oj 1236 求最小公倍数( lcm(a,b) )等于n的数对素因数分解

H - Pairs Forming LCM

Time Limit:2000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1236

Description

Find the result of the following code:

long long pairsFormLCM( int n ) {
  long long res = 0;
  for( int i = 1; i <= n; i++ )
  for( int j = i; j <= n; j++ )
if( lcm(i, j) == n ) res++; // lcm means least common multiple
  return res;
}

A straight forward implementation of the code may time out. If you analyze the code, you will find that the code actually counts the number of pairs (i, j) for which lcm(i, j) = n and (i ≤ j).

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 10¹⁴).

Output

For each case, print the case number and the value returned by the function 'pairsFormLCM(n)'.

Sample Input

Sample Output

Case 1: 2

Case 2: 2

Case 3: 3

Case 4: 5

Case 5: 4

Case 6: 5

Case 7: 8

Case 8: 5

Case 9: 8

Case 10: 8

Case 11: 5

Case 12: 11

Case 13: 3

Case 14: 4

Case 15: 2

题意：对给定的n，求符合lcm(a,b)==n的(a,b)的数对的个数。

思路：对n进行素数分解，a和b与n对应分解

n=p1^e1*p2^e2*...*pk^ek;

a=p1^a1*p2^a2*…*pk^ak
b=p1^b1*p2^b2*…*pk^bk
对于a,b它们的最小公倍数为n,则a1<=e1,b1<=e1,则(a1,b1)有2*(e1+1)-1对则总的对数有t=(2*e1+1)*(2*e2+1)....，其中除了（n,n）,其他的都重复了两次则答案为(t+1)/2.

注意由于n的范围是10^14，因此素数筛到10^7，然而还是会漏过素因子，但显然只会漏过最多一个素因子，因为sqrt(10^14)==10^7，如果有两个大于10^7的素因子，根据素数分解公式，必定含有这两个大于10^7的素因子，相乘一定大于10^14大于n，与分解等式矛盾。因此只需最后特判最后一个即可。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<string>
#include<math.h>
#include<cctype>using namespace std;typedef long long ll;
const int maxn=10000100;
const int INF=(1<<29);
const double EPS=0.0000000001;
const double Pi=acos(-1.0);int T;
ll n;
vector<ll> prime;
bool isprime[maxn];void play_prime()
{memset(isprime,1,sizeof(isprime));isprime[1]=0;for(int i=2;i<maxn;i++){if(!isprime[i]) continue;for(int j=i*2;j<maxn;j+=i) isprime[j]=0;}for(int i=1;i<maxn;i++) if(isprime[i]) prime.push_back(i);
}int main()
{int tag=1;cin>>T;play_prime();while(T--){cin>>n;ll ans=1;for(int i=0;i<prime.size()&&prime[i]<=n;i++){ll t=prime[i];int cnt=0;while(n%t==0){n/=t;cnt++;}ans*=(2*cnt+1);}if(n>1) ans*=(2*1+1);ans=(ans+1)/2;printf("Case %d: %lld\n",tag++,ans);}return 0;
}

View Code

转载于:https://www.cnblogs.com/--560/p/4552948.html