Light OJ 1278 Sum of Consecutive Integers N拆分成连续整数和

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题目来源：Light OJ 1278 Sum of Consecutive Integers

题意：N拆分成连续整数和的方案数

思路：奇因数的个数

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
//筛素数 
const int maxn = 10000010;
bool vis[maxn];
int prime[1000000];

int sieve(int n)
{
	memset(vis, 0, sizeof(vis));
	vis[0] = vis[1] = 1;
	int c = 0;
	for(int i = 2; i <= n; i++)
		if(!vis[i])
		{
			prime[c++] = i;
			for(int j = 2*i; j <= n; j += i)
				vis[j] = 1;
		}
	return c;
}


int main()
{
	int c = sieve(10000000);
	int cas = 1;
	int T;
	scanf("%d", &T);
	while(T--)
	{
		long long n, ans = 1;
		scanf("%lld", &n);
		while(n%2 == 0)
			n /= 2;
		for(int i = 0; i < c && (long long)prime[i]*prime[i] <= n; i++)
		{
			if(prime[i] > n)
				break;
			if(n % prime[i] == 0)
			{
				long long sum = 1;
				while(n % prime[i] == 0)
				{
					sum++;
					n /= prime[i];
				}
				ans *= sum;
			}
		}
		if(n > 1 && (n&1))
			ans *= 2;
		printf("Case %d: %lld\n", cas++, ans-1);
	}
	return 0;
}

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Light OJ 1278 Sum of Consecutive Integers N拆分成连续整数和

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原文地址：http://blog.csdn.net/u011686226/article/details/28672983