Description
Given A,B,C, You should quickly calculate the result of A^B mod C. (1<=A,C<=1000000000,1<=B<=10^1000000).
Input
There are multiply testcases. Each testcase, there is one line contains three integers A, B and C, separated by a single space.
Output
For each testcase, output an integer, denotes the result of A^B mod C.
Sample Input
3 2 4
2 10 1000
Sample Output
1
24
首先要降幂，由降幂公式：
（然而不知道公式是怎么来的。。。求指教）降幂后用快速幂计算
1 #include<iostream> 2 #include<cstdio> 3 #include<algorithm> 4 #include<cstring> 5 #include<cmath> 6 #define ll __int64 7 #define N 1000100 8 using namespace std; 9 char b[N]; 10 ll p[N]; 11 ll a, c; 12 ll quick(ll a, ll b){ //快速幂 13 ll k = 1; 14 while(b){ 15 if(b%2==1){ 16 k = k*a; 17 k %=c; 18 } 19 a = a*a%c; 20 b /=2; 21 } 22 return k; 23 } 24 void priem(){ 25 memset(p, 0, sizeof(p)); 26 ll i, j; 27 p[1] = 1; 28 for(i=2; i<=sqrt(N); i++){ 29 for(j=2; j<=N/i; j++) 30 p[i*j] = 1; 31 } 32 } 33 ll ola(ll n){ //欧拉函数 34 ll i, j, r, aa; 35 r = n; 36 aa = n; 37 for(i=2; i<=sqrt(n); i++){ 38 if(!p[i]){ 39 if(aa%i==0){ 40 r = r/i*(i-1); 41 while(aa%i==0) 42 aa /= i; 43 } 44 } 45 } 46 if(aa>1) 47 r = r/aa*(aa-1); 48 return r; 49 } 50 int main(){ 51 ll d, i, j; 52 priem(); 53 while(~scanf("%I64d%s%I64d",&a,b,&c)){ 54 ll l = strlen(b); 55 ll B=0; 56 ll oc = ola(c); 57 // cout<<"oc = "<<oc<<endl; 58 for(i=0; i<l; i++){ 59 B = B*10+b[i]-'0'; 60 if(B>oc) 61 break; 62 } 63 //cout<<i<<endl; 64 if(i==l) 65 d = quick(a,B); 66 else{ 67 B=0; 68 for(i=0; i<l; i++){ //降幂 69 B = (B*10+b[i]-'0')%oc; 70 } 71 d = quick(a,B+oc); 72 } 73 // printf("B= %I64d ",B); 74 printf("%I64d ",d); 75 } 76 return 0; 77 }