FOJ Problem 1759 Super A^B mod C

2011-04-20  20:04:55

题目地址：http://acm.fzu.edu.cn/problem.php?pid=1759

公式：A^B mod C =  A^(B%euler(C))+euler(C)) mod C

证明：http://hi.baidu.com/aekdycoin/blog/item/b6f1762565bb403fc8955908.html

 

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#include<iostream> #include<cmath> #define N 100000000000000000 #define ui64 unsigned __int64 using namespace std; char str[1000001]; ui64 euler(ui64 n) { ui64 euler=n,nn=n,i; if(n==1) return 1; for(i=2;i*i<=n;i++) { if(nn%i==0) { euler=euler-euler/i; while(nn%i==0) nn/=i; } if(nn==1) break; } if(n==nn) euler=n-1; if(nn!=n&&nn!=1) euler=euler-euler/nn; return euler; } ui64 mul(ui64 a,ui64 b, ui64 n) { ui64 ret=0; a%=n; while (b) { if(b&1){ret+=a;} if(ret>n) ret-=n; a<<=1; if(a>n) a-=n; b>>=1; } return ret; } ui64 mod(ui64 x,ui64 e,ui64 n) { ui64 ret=1; x%=n; while (e) { if(e&1) {ret=mul(ret,x,n);} x=mul(x,x,n); e>>=1; } return ret; } int main() { int i; ui64 b,a,c,eul; while (scanf("%I64u%s%I64u",&a,str,&c)!=EOF) { eul=euler(c); b=0; for(i=0;str[i]!='/0';) { while(b<N&&str[i]!='/0') b=b*10+str[i++]-'0'; b%=eul; } b+=eul; printf("%I64u/n",mod(a,b,c)); } return 0; }