Uva10290 - {Sum+=i++} to Reach N

Problem H

{sum+=i++} to Reach N

Input: standard input

Output:  standard output

Memory Limit: 32 MB

All the positive numbers can be expressed as a sum of one, two or more consecutive positive integers. For example 9 can be expressed in three such ways, 2+3+4, 4+5 or 9. Given an integer
less than (9*10^14+1) or (9E14 + 1) or (9*10¹⁴ +1) you will have to determine in how many ways that number can be expressed as summation of consecutive numbers.

 

Input

The input file contains less than 1100 lines of input. Each line contains a single integer N  (0<=N<= 9E14). Input is terminated by end of file.

 

Output

For each line of input produce one line of output. This line contains an integer which tells in how many ways N can be expressed as summation of consecutive integers.

Sample Input

9

11

12

 

Sample Output

3

2

2

题意：问你N能够由多少种方案：连续的x个整数相加和为N

思路：转化为求奇因数，分解质因数后求排列数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 1e7+10;
typedef long long ll;

bool isPrime[maxn];
vector<int> prime,cnt;
ll n;

void getPrime(){
    memset(isPrime,1,sizeof isPrime);
    for(int i = 2; i < maxn; i++){
        if(isPrime[i]){
            prime.push_back(i);
            for(int j = i+i; j < maxn; j+=i){
                isPrime[j] = 0;
            }
        }
    }
}

void getDigit(){

    while(n%2==0) n/=2;
   // cout<<n<<endl;
    for(int i = 1; i < prime.size()&&n >= prime[i]; i++){
        if(n%prime[i]==0){
            int t = 0;
            while(n%prime[i]==0){
                n /= prime[i];
                t++;
            }
            cnt.push_back(t);
        }
    }
    if(n!=1) cnt.push_back(1);
}

int main(){

    getPrime();
    while(~scanf("%lld",&n)){
        cnt.clear();
        getDigit();
        ll ans = 1;
        for(int i = 0; i < cnt.size(); i++) ans *= (cnt[i]+1);
        cout<<ans<<endl;

    }
    return 0;
}