2016 Multi-University Training Contest 1 F.PowMod
PowMod
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1008 Accepted Submission(s): 341
Problem Description
Declare:
k=∑mi=1φ(i∗n) mod 1000000007
n is a square-free number.
φ is the Euler's totient function.
find:
ans=kkkk...k mod p
There are infinite number of k
k=∑mi=1φ(i∗n) mod 1000000007
n is a square-free number.
φ is the Euler's totient function.
find:
ans=kkkk...k mod p
There are infinite number of k
Input
Multiple test cases(test cases ≤100), one line per case.
Each line contains three integers, n,m and p.
1≤n,m,p≤107
Each line contains three integers, n,m and p.
1≤n,m,p≤107
Output
For each case, output a single line with one integer, ans.
Sample Input
1 2 6
1 100 9
Sample Output
4
7
Author
HIT
Source
题意: 令K = sigma(phi(i * n)) 1 <= i <= m 求K的超级幂。
题解: 关于下面要用到的欧拉函数的几个证明。 1、首先显然phi(p) = p-1, phi(1) = 1 2、若 n = p^k,则 phi(n) = p^k - p^(k-1) 证明:显然不互质的有p,2p,......,(p^(k-1)-1)*p,以及p^k 所以共有p^(k-1)个 3、若 n = a * b, gcd(a, b) == 1, 则phi(n) = phi(a) * phi(b) 证明: phi(a)个与a互质的数,设它们为A1,A2......Aphi(a) 同理,我们也有B1,B2,.....,Bphi(b)为与b互质的数。 那么与n互质的数为Ai*b+Bj*a,共有phi(a)*phi(b)个。 我们来考察Ai*b+Bj*a的性质。 我们要证明两点: 1、Ai*b+Bj*a与n互质 2、除了这些数,没有与n互质的数了。 第一点、如果有gcd(a*Bi+b*Ai,a*b)=x>1 那么x|a*b,又gcd(a,b)=1,所以必有x|a或者x|b且不同时成立。 不妨假设x|a,那么a*Bi/x是个整数,而gcd(Ai,a)=1,所以gcd(Ai,x)=gcd(b,x)=1, 喜爱内燃此时b*Ai/x不为整数,与x为最大公约数的假设矛盾。 第二条、这是显然的。。。如果有gcd(x,a*b)!=1,那么gcd(x,a)!=1和gcd(x,b)!=1至少有一条成立。 所以证毕。 4、若n=p1^k1*p2&k2*.....*pm^km,那么 phi(n)=n*(1-1/p1)*(1-1/p2)*.....*(1-1/pm) 5、当n>2时,phi(n)是偶数,因为1-1/p=(p-1)/p,p-1必然是偶数。 K = sigma(phi(i * n)) % Q, Q = 1e9+7, 1 <= i <= m = sigma(phi( (i * n) / p * p )), p|n = sigma(phi( i * n / p * p)), i % p != 0 + sigma(phi( i * n * p)), 1 <= i <= floor(m / p) % Q = sigma(phi( i * n / p) * phi(p)), i%p != 0 + sigma(phi(i * (n/p) * p^2)), 1 <= i <= floor(m / p), %Q = phi(p) * sigma(phi(i * n / p)), i%p != 0 + sigma(phi(i * (n / p)) * phi(p^2)), 1 <= i <= floor(m/p) %Q = phi(p) * sigma(phi(i * n / p)) , i % p != 0 + sigma(phi(i * (n / p)) * p * phi(p)), 1 <= i <= floor(m/p) %Q = phi(p) * sigma(phi(i * n / p)) , i % p != 0 + sigma(phi(i * (n / p)) * (phi(p) + 1) * phi(p)), 1 <= i <= floor(m/p) %Q = phi(p) * ( sigma(phi(i * n / p)), i % p != 0 + sigma(phi(i * n)), 1<=i<=floor(m/p) ) + sigma(phi(i * n)), 1<=i<=floor(m/p) %Q = phi(p) * sigma(phi(i * (n / p))), 1 <= i <= m + sigma(phi(i * n)), 1 <= i <= floor(m / p) 令F(m,n) = sigma(phi(i * n)) % Q, 根据上述证明,有当p|n时, F(m, n) = phi(p) * F(m, n / p) + F(floor(m / p), n) %Q 所以可以递归计算F(m, n), 这一步复杂度为O((number of P)^2) 当算出K值后,需要计算它的超级幂。 有A^B mod C = A^(B % phi(C) + phi(C)) % C 这个证明我不会,网上有证明,但因为年代久远,百度博客搬迁,原地址没了。。。 (实际上这个的作用除了计算超级幂好像没什么用。。。。(这只是弱鸡的视野 事实上,这里的A跟B是一个东西,而phi(C)是不断减少的,只有phi(1)==1, 而且收敛速度很快。 所以不断递归就好啦。
1 const int N = 10000010, MOD = 1e9 + 7; 2 int prime[N], tot; 3 bool notPrime[N]; 4 int phi[N], sumphi[N]; 5 6 int n, m, p; 7 int factor[N], totFactor; 8 9 inline int add(int x, int y, int MOD = MOD) { 10 return ((x + y) % MOD + MOD) % MOD; 11 } 12 13 inline int mul(int x, int y, int MOD = MOD) { 14 return ((x * 1ll * y) % MOD + MOD) % MOD; 15 } 16 17 inline void getPrime() { 18 phi[1] = 1, sumphi[1] = 1; 19 for(int i = 2; i < N; ++i) { 20 if(!notPrime[i]) prime[tot++] = i, phi[i] = i - 1; 21 for(int j = 0; j < tot; ++j) { 22 if(i * prime[j] >= N) break; 23 notPrime[i * prime[j]] = true; 24 if(i % prime[j]) phi[i * prime[j]] = phi[i] * phi[prime[j]]; 25 else { 26 phi[i * prime[j]] = phi[i] * prime[j]; 27 break; 28 } 29 } 30 sumphi[i] = add(phi[i], sumphi[i - 1]); 31 } 32 } 33 34 inline void getFactor(int x) { 35 totFactor = 0; 36 for(int i = 0; i < tot; ++i) 37 if(!(x % prime[i])) { 38 factor[totFactor++] = prime[i]; 39 x /= prime[i]; 40 if(x <= 1) break; 41 } 42 } 43 44 inline int f(int index, int m, int n) { 45 if(index >= totFactor || n == 1) return sumphi[m]; // n == 1 46 if(m <= 0) return 0; 47 int part1 = f(index + 1, m, n / factor[index]), 48 part2 = f(index, m / factor[index], n); 49 return add(mul(phi[factor[index]], part1), part2); 50 } 51 52 inline int fastpow(int basic, int times, int p) { 53 int ret = 1; 54 while(times) { 55 if(times & 1) ret = mul(ret, basic, p); 56 basic = mul(basic, basic, p), times >>= 1; 57 } 58 return ret; 59 } 60 61 inline int superPower(int k, int p) { 62 /** 63 * 1. k^X mod p = k^(X mod phi(p) + phi(p)) mod p X = k^k^k..... 64 * 2. fastpow(x, y, z) x^y mod z 65 * */ 66 if(p == 1) return 0; 67 int powers = superPower(k, phi[p]) + phi[p]; 68 return fastpow(k, powers, p); 69 } 70 71 inline void solve() { 72 getFactor(n); 73 int k = f(0, m, n); 74 int ans = superPower(k, p); 75 printf("%d\n", ans); 76 } 77 78 int main() { 79 getPrime(); 80 while(scanf("%d%d%d", &n, &m, &p) == 3) solve(); 81 return 0; 82 }