例题10-4 最小公倍数的最小和 UVa10791

标签：uva   数论   

1.题目描述：点击打开链接

2.解题思路：本题要求找至少两个整数，使得它们的最小公倍数是n。本题看似简单，但还是应该注意细节，考虑周密。当n=1时答案是2，当n只有一种素因子时答案是n+1，由于n的最大范围是2^31-1，因此保险起见用long long防止溢出。

3.代码：

#define _CRT_SECURE_NO_WARNINGS 
#include<iostream>
#include<algorithm>
#include<string>
#include<sstream>
#include<set>
#include<vector>
#include<stack>
#include<map>
#include<queue>
#include<deque>
#include<cstdlib>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<ctime>
#include<functional>
using namespace std;

typedef long long LL;
const int maxn = 1000000;
int vis[maxn];
int e[maxn];
vector<LL>primes;
void init()
{
	int m = sqrt(maxn + 0.5);
	for (int i = 2; i <= m;i++)
	if (!vis[i])
	for (int j = i*i; j <= maxn; j+=i)
		vis[j] = 1;
	for (int i = 2; i <= maxn;i++)
	if (!vis[i])
		primes.push_back(i);
}
bool is_prime(LL n)
{
	if (n <= maxn)
	{
		if (!vis[n])return true;
		else return false;
	}
	int m = sqrt((double)n + 0.5);
	for (int i = 0; i < primes.size(); i++)
	{
		if (primes[i]>m)break;
		if (n%primes[i] == 0)return false;
	}
	return true;
}

void solve(LL n)
{
	LL tmp = n;
	int cnt = 0;
	for (int i = 0; i < primes.size(); i++)
	{
		int ok = 0;
		while (tmp%primes[i] == 0)
		{
			ok = 1;
			tmp /= primes[i];
			e[i]++;
		}
		if (ok)cnt++;
	}
	if (cnt == 1)cout << n + 1 << endl;
	else if (cnt > 1)
	{
		LL sum = 0;
		for (int i = 0; i <= maxn;i++)
		if (e[i] > 0)
			sum += pow(primes[i], e[i]);
		cout << sum << endl;
	}
}
int main()
{
	//freopen("test.txt", "r", stdin);
	LL n;
	int rnd = 0;
	init();
	while (scanf("%lld", &n) != EOF&&n)
	{
		memset(e, 0, sizeof(e));
		printf("Case %d: ", ++rnd);
		if (n == 1)cout << 2 << endl;
		else
		{
			if (n <= maxn&&!vis[n])
				cout << n + 1 << endl;
			bool ok = is_prime(n);
			if (n > maxn&&ok)
				cout << n + 1 << endl;
			if (!ok)
				solve(n);
		}
	}
	return 0;
}

例题10-4 最小公倍数的最小和 UVa10791

标签：uva   数论   

原文地址：http://blog.csdn.net/u014800748/article/details/43890289