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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
15
2
3
4
6
8
10
12
15
18
20
21
24
25
27
29
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
#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; }
light_oj 1236 求最小公倍数( lcm(a,b) )等于n的数对 素因数分解
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原文地址：http://www.cnblogs.com/--560/p/4552948.html
