codeforces100917 Constant Ratio

codeforces100917 Constant Ratio 题目链接：http://codeforces.com/gym/100917/problem/C

题面描述：

C. Constant Ratio

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Given an integer n, find out number of ways to represent it as the sum of two or more integers ai with the next property: ratio ai / ai - 1 is the same positive integer for all possible i > 1.

Input

Input consists of one integer n (1 ≤ n ≤ 105).

Output

Print one integer — number of representations.

Examples

input

1

output

0

input

5

output

2

input

567

output

21

Note

In the first sample no such representation exists.

In the second sample there exist two representations:

• 1 1 1 1 1, then q = 1.
• 1 4, then q = 4.

题目大意：

求等比数列的前n项和，看前n项和能否得到n。

题目分析：

可以先找出每一个可以被n整除的数，作为a1，然后再找公比，由于是等比数列，所以时间复杂度不是很高。

代码实现：方法一：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <cstdlib>#include <cmath>using namespace std;typedef long long LL;int n;bool find(int a1,int a2){    int k=a2/a1;    LL ak=a1;    LL sum=a1;    while(1)    {        ak*=k;        sum+=ak;        if (sum==n) return true;        if (sum>n) return false;    }}int main(){    while(scanf("%d",&n)!=-1)    {        int ans=0;        for(int i=1; i<n; i++) if (n%i==0)        {            for(int j=i; j<=n; j+=i)            {                if (find(i,j)) ans++;            }        }        printf("%d\n",ans);    }    return 0;}

方法二：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>using namespace std;long long n;long long pow(long long a,long long q){    long long ans=1;    while(q--)        ans*=a;    return ans;}bool find(long long a1,long long a2){    long long q=a2/a1;    if(q==1)    {        if(n%a1==0)            return true;        else            return false;    }    for(long long k=1; k<=n*n; k++)///注意k的范围给的给定,要使得能够跳出循环    {        if(n==(a1*(pow(q,k)-1)/((q-1))))            return true;        else if((a1*(pow(q,k)-1)/((q-1)))>n)            return false;        else continue;    }}int main(){    while(scanf("%lld",&n)!=-1)    {        if(n==1)            printf("0\n");        else        {            long long ans=0;            for(long long i=1; i<n; i++)//a1,不带n            {                if(n%i==0)                {                    for(long long j=i; j<=n; j+=i)                    {                        if(find(i,j)) ans++;                    }                }            }            printf("%lld\n",ans);        }    }    return 0;}