Sum of Consecutive Integers LightOJ - 1278————（算术基本定理+容斥）

Given an integer $N$, you have to find the number of ways you can express $N$ as sum of consecutive integers. You have to use at least two integers.

For example, $N = 15 \;$has three solutions, $(1+2+3+4+5), (4+5+6), (7+8).$

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

Each case starts with a line containing an integer $N (1 ≤ N ≤ 10^{14}).$

Output
For each case, print the case number and the number of ways to express$\;N$ as sum of consecutive integers.

Sample Input
5
10
15
12
36
828495
Sample Output
Case 1: 1
Case 2: 3
Case 3: 1
Case 4: 2
Case 5: 47

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给你一个正整数$\ N$ ，用一段连续正整数的和表示$N$的方法有多少种？

也就是找奇数因子的个数

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#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
using  namespace std;
typedef long long LL;
const int maxn = 10000000, INF = 0x7fffffff;
int primes[1000000];
bool vis[maxn];
int ans = 0;
void init()
{
    memset(vis,0,sizeof(vis));
    for(int i=2; i<maxn; i++)
    {
        if(vis[i]) continue;
        primes[ans++] = i;
        for(LL j=(LL)i*i; j<maxn; j+=i)
            vis[j] = 1;
    }
}
int main()
{
    init();
    int t;
    scanf("%d",&t);
    int kase=0;
    while(t--)
    {
         LL n, res = 1;
        cin>> n;
        for(int i=0; i<ans && primes[i]*primes[i] <= n; i++)
        {
            LL cnt2 = 0;
            while(n % primes[i] == 0)
            {
                n /= primes[i];
                cnt2++;
            }
            if(cnt2 > 0 && primes[i] % 2)
                res *= (cnt2 + 1);
        }
        if(n > 1 && n % 2)
        {
            res *= 2;
        }
        printf("Case %d: %lld\n", ++kase, res - 1);
    }
    return 0;
}