这题模数是9937还不是素数,求逆元还得手动求。
项链翻转一样的算一种相当于就是一种类型的置换,那么在n长度内,对于每个i其循环节数为(i,n),但是由于n<=2^32,肯定不能直接枚举,所有考虑枚举gcd,对应的n/gcd就是其个数,有点容斥的思想。全部累加最后除以n就计算好染色方案了。
注意这题很卡时间,而且很玄的用long long会错,要先求出上界再枚举,循环中i*i的循环条件会很慢。
/** @Date : 2017-09-18 23:33:30
* @FileName: HDU 2239 EXGCD 求逆元.cpp
* @Platform: Windows
* @Author : Lweleth (SoungEarlf@gmail.com)
* @Link : https://github.com/
* @Version : $Id$
*/
#include <bits/stdc++.h>
#define LL __int64
#define PII pair
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std; const int INF = 0x3f3f3f3f;
const int N = 1e5 + 10;
const double eps = 1e-8;
const LL mod = 9937;
int pri[N];
int c = 0;
bool vis[N];
void prime()
{
MMF(vis);
for(int i = 2; i < N; i++)
{
if(!vis[i]) pri[c++] = i;
for(int j = 0; j < c && i * pri[j] < N; j++)
{
vis[i *pri[j]] = 1;
if(i % pri[j] == 0)
break;
}
}
} LL get_phi(int x)
{
LL res = x;
int ma = sqrt(x + 0.5);
for(int i = 0; i < c && pri[i] <= ma; i++)
{
if(x % pri[i] == 0)
{
while(x % pri[i] == 0)
x /= pri[i];
res = res / pri[i] * (pri[i] - 1);
}
}
if(x > 1) res = res / x * (x - 1);
return res % mod;
}
LL exgcd(LL a, LL b, LL x, LL y)
{
LL d = a;
if(b == 0)
x = 1, y = 0;
else
{
d = exgcd(b, a % b, y, x);
y -= (a / b) * x;
}
return d;
} LL fpow(LL a, LL n)
{
LL res = 1;
while(n)
{
if(n & 1)
res = res * a % mod;
a = a * a % mod;
n >>= 1;
}
return res;
} int main()
{
LL n, m;
prime(); while(~scanf("%I64d%I64d", &n, &m))
{
LL ans = 0;
int ma = sqrt(n + 0.5);
for(int i = 1; i <= ma; i++)
{
if(n % i != 0)
continue;
ans = (ans + get_phi(n / i) * fpow(m, i) % mod + mod) % mod;
if(i != n / i)
ans = (ans + get_phi(i) * fpow(m, n / i) % mod + mod) % mod;
}
for(LL i = 0; i < mod; i++)
if(i * n % mod == ans % mod)
{
ans = i;
break;
}
/*cout << ans << endl;
ans = ans * mod % (n * mod) / n;*/
printf("%I64d\n", ans);
}
return 0;
}