【FZU - 1759】Super A^B mod C (數論,快速冪,快速乘,尤拉降冪,指數迴圈節,模板)
阿新 • • 發佈:2018-11-01
題幹:
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
解題報告:
裸題,僅供模板準備。
AC程式碼:
#include <iostream> #include <string.h> #include <stdio.h> using namespace std; const int N=1000005; typedef long long LL; char str[N]; int phi(int n) { int rea = n; for(int i=2; i*i<=n; i++) { if(n % i == 0) { rea = rea - rea / i; while(n % i == 0) n /= i; } } if(n > 1) rea = rea - rea / n; return rea; } LL multi(LL a,LL b,LL m) { LL ans = 0; a %= m; while(b) { if(b & 1) { ans = (ans + a) % m; b--; } b >>= 1; a = (a + a) % m; } return ans; } LL quick_mod(LL a,LL b,LL m) { LL ans = 1; a %= m; while(b) { if(b & 1) { ans = multi(ans,a,m); b--; } b >>= 1; a = multi(a,a,m); } return ans; } void Solve(LL a,char str[],LL c) { LL len = strlen(str); LL ans = 0; LL p = phi(c); if(len <= 15) { for(int i=0; i<len; i++) ans = ans * 10 + str[i] - '0'; } else { for(int i=0; i<len; i++) { ans = ans * 10 + str[i] - '0'; ans %= p; } ans += p; } printf("%I64d\n",quick_mod(a,ans,c)); } int main() { LL a,c; while(~scanf("%I64d%s%I64d",&a,str,&c)) Solve(a,str,c); return 0; }