Gcd HDU - 6545 (basic number theory)

wls have an integer n, he wanted to 1 - n n digital this divided into two groups, each group having at least one number, and such that the greatest common divisor of two numbers and the maximum, output a maximum common divisor.
Input
input line of an integer n.
. 1 ≤ ≤ n-2, 000, 000, 000
the Output
output line answers an integer.
The Input the Sample
6
the Sample the Output
7

Ideas:

We pray 1 to n and the sum for the sum,

And the sum is divided into two groups respectively sum1 and sum2,

Then the meaning of the questions we know sum1 + sum2 = sum

The answer is in demand gcd (sum1, sum2) maximum, we set ans

We know that according to the nature of gcd gcd (sum1, sum2) = gcd (sum1, sum) = gcd (sum2, sum)

Because gcd (x, y) = gcd (x, x + y)

So we can draw the equation:

sum1 / year + sum2 / year = sum / year

Ans is obviously a factor sum, we think is the biggest ans, that is the biggest factor sum.

How to find a number of the biggest factors it? We just need to find the sum violent enumerate the smallest factor x, sum / x = ans.

See details Code:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define chu(x) cout<<"["<<#x<<" "<<(x)<<"]"<<endl
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/

int main()
{
    //freopen("D:\\common_text\\code_stream\\in.txt","r",stdin);
    //freopen("D:\\common_text\\code_stream\\out.txt","w",stdout);
    
    
    ll n;
    cin>>n;
    if(n==2)
    {
        return puts("1");
    }else
    {
        ll sum=(n+1ll)*n/2ll;
        for(ll i=2ll;;++i)
        {
            if(sum%i==0)
            {
                cout<<sum/i<<endl;
                break;
            }
        }
    }
    
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}