LightOJ1236 Pairs Forming LCM
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题目链接:点我
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
152346810121518202124252729
Sample Output
Case 1: 2Case 2: 2Case 3: 3Case 4: 5Case 5: 4Case 6: 5Case 7: 8Case 8: 5Case 9: 8Case 10: 8Case 11: 5Case 12: 11Case 13: 3Case 14: 4Case 15: 2
题意:
求前n个数组合的最小公倍数等于n的数有多少组.
思路:
由唯一分解定理我们知道,n =
假若指数e出现在a中,那么b中的指数有 e + 1 种情况,假若没有出现在a中,那么a有 e 种情况,而b只能取e,所有共有2 * e + 1 种情况,由于题目中要求a <= b,所以我们将答案除以2, 然后加上 a = b 的一种情况.
代码:
#include<cstring>#include<algorithm>#include<cstdio>#include<cmath>#include<iostream>using namespace std;const int maxn = 1e7 + 10;typedef long long LL;bool isprime[maxn];int pri[maxn/10];int num;void prime(){ memset(isprime,false,sizeof(isprime)); num = 0; for(int i = 2; i < maxn; ++i){ if(!isprime[i]) pri[++num] = i; for(int j = 1; j <= num && i * pri[j] < maxn; ++j){ isprime[i * pri[j]] = true; if( i % pri[j] == 0) break; } }}int main(){ LL n; int t; prime(); int kase = 0; scanf("%d", &t); while(t-- ){ scanf("%lld", &n); LL ans = 1; for(int i = 1; i <= num && pri[i] <= (int)sqrt(n); ++i){ LL cnt = 0; while(n % pri[i] == 0){ n /= pri[i]; ++cnt; } ans *= (2 * cnt + 1) ; } if(n > 1) ans *= 3; printf("Case %d: %lld\n",++kase, ans / 2 + 1); } return 0;}
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