1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<cmath> 5 #define ll long long 6 #define mul 999911659 7 using namespace std; 8 int n,g,a[4]; 9 int sh[4]={2,3,4679,35617},C1[4][35618]; 10 void exgcd(int a1,int a2,int &x,int &y) 11 { 12 if(!a2) 13 { 14 x=1; 15 y=0; 16 return; 17 } 18 exgcd(a2,a1%a2,x,y); 19 int t=x; 20 x=y; 21 y=t-a1/a2*y; 22 } 23 int kuai(ll n,int k,int p) 24 { 25 int ans=1; 26 for(;k;) 27 { 28 if(k%2) 29 ans=(ans*n)%p; 30 n=(n*n)%p; 31 k/=2; 32 } 33 return ans; 34 } 35 int C(int n,int m,int p) 36 { 37 if(n<m) 38 return 0; 39 return C1[p][n]*kuai(C1[p][m]*C1[p][n-m],sh[p]-2,sh[p])%sh[p]; 40 } 41 int lucas(int n,int m,int p) 42 { 43 if(!m) 44 return 1; 45 return (C(n%sh[p],m%sh[p],p)*lucas(n/sh[p],m/sh[p],p))%sh[p]; 46 } 47 int solve() 48 { 49 int x,y,a1,b1,a2,b2; 50 a1=sh[0]; 51 b1=a[0]; 52 for(int i=1;i<4;i++) 53 { 54 a2=sh[i]; 55 b2=a[i]; 56 exgcd(a1,a2,x,y); 57 x=((b2-b1)*x%a2+a2)%a2; 58 b1=b1+x*a1; 59 a1=a1*a2; 60 } 61 return b1; 62 } 63 int main() 64 { 65 scanf("%d%d",&n,&g); 66 if(g==mul) 67 { 68 printf("0\n"); 69 return 0; 70 } 71 for(int i=0;i<4;i++) 72 { 73 C1[i][0]=1; 74 for(int j=1;j<=sh[i];j++) 75 C1[i][j]=(C1[i][j-1]*j)%sh[i]; 76 } 77 for(int i=1;i<=sqrt(n);i++) 78 if(n%i==0) 79 { 80 for(int j=0;j<4;j++) 81 { 82 a[j]=(a[j]+lucas(n,i,j))%sh[j]; 83 if(i!=n/i) 84 a[j]=(a[j]+lucas(n,n/i,j))%sh[j]; 85 } 86 } 87 printf("%d\n",kuai(g,solve(),mul)); 88 return 0; 89 }
经典的数学题。。。。
题目就有点难懂，求G^M mod P M=∑ i|N C(N,i) P=999911659
用lucas定理，中国剩余定理合并模线性方程组。http://hzwer.com/4407.html
