poj 1286

题目描述：

题解：

重点：

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <ctype.h>
#include <limits.h>
#include <cstdlib>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#include <set>
#include <bitset>
#define CLR(a) memset(a, 0, sizeof(a))
#define REP(i, a, b) for(ll i = a;i < b;i++)
#define REP_D(i, a, b) for(ll i = a;i <= b;i++)

typedef long long ll;

using namespace std;

const ll maxn = 30 + 10;
ll p[maxn];
ll n;

ll gcd(ll a, ll b)
{
if(b==0)
return a;
return gcd(b, a%b);
}
void getP()
{
ll key = 24;
p[0] = 1;
for(ll i = 1;i < key;i++)
{
p[i] = p[i-1]*3;
}
}
void solve()
{
ll ans = 0;
ans += p
;
for(ll i = 1;i < n;i++)
{
ll t = gcd(i, n);
ans += p[t];
}
if(n%2==0)
{
ans += (n/2)*(p[n/2]) + (n/2)*(p[n/2+1]);
}
else
{
ans += (n)*(p[n/2+1]);
}
ans /= 2*n;
printf("%I64d\n", ans);
}
int main()
{
//freopen("7Gin.txt", "r", stdin);
//freopen("7Gout.txt", "w", stdout);
getP();
while(scanf("%I64d", &n) != EOF)
{
if(n==-1)
break;
if(n==0)
printf("0\n");
else
solve();
}
return 0;
}