a^b%c问题 ---模板

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(1) ABmod C.(1<=A,B<2^62, 1 <= C <= 10^9)

http://acm.bit.edu.cn/mod/programming/view.php?a=530

快速幂----二分

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>using namespace std;long long quickpow(long long n,long long m,long long p){long long t=1;while(m){if(m&1) t=(t*n)%p;m=m>>1;n=(n*n)%p;}return t;}int main(){long long a,b,c;while(~scanf("%lld%lld%lld",&a,&b,&c)){a%=c;    //a要先做处理printf("%lld\n",quickpow(a,b,c));}}

(2)ABmod C.(1<=A,B,C<2^62)

http://acm.bit.edu.cn/mod/programming/view.php?id=696

c的范围变为 long long ,a%=c后,a的范围还是long long.当二分乘时,a*a会超long long,

这是,我们把两个数相乘a*b也用二分算,a*b==(a*2) * (b/2);

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>using namespace std;long long cheng(long long a,long long b,long long p){long long t=0;while(b){if(b&1) t=(t+a)%p;b=b>>1;a=(a+a)%p;}return t;}long long quickpow(long long n,long long m,long long p){long long t=1;while(m){if(m&1) t=cheng(t,n,p);m=m>>1;n=cheng(n,n,p);}return t;}int main(){long long a,b,c;while(~scanf("%lld%lld%lld",&a,&b,&c)){a%=c;printf("%lld\n",quickpow(a,b,c));}}

(3)  A^B mod C. (1<=A,C<=1000000000,1<=B<=10^1000000).

http://acm.bit.edu.cn/mod/programming/view.php?a=542

b是高精度,

有公式:

公式具体内容:http://hi.baidu.com/aekdycoin/item/e493adc9a7c0870bad092fd9

对b进行处理,讲b处理为  int 型数。然后用快速幂就可以了,

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<algorithm>using namespace std;long long quickpow(long long n,long long m,long long p){long long t=1;while(m){if(m&1) t=t*n%p;m=m>>1;n=n*n%p;}return t;}long long eular(long long n){      long long ret=1,i;      for (i=2;i*i<=n;i++)          if (n%i==0){              n/=i,ret*=i-1;              while (n%i==0)                  n/=i,ret*=i;          }      if (n>1)          ret*=n-1;      return ret;  }char b[1000005];int main(){long long a,c;while(~scanf("%lld%s%lld",&a,b,&c)){a%=c;long long tb=0;long long phic=eular(c);int flag=0;int len=strlen(b);for(int i=0;i<len;i++){tb=tb*10+b[i]-'0';if(tb>=phic) tb%=phic,flag=1;}if(flag==1) tb+=phic;printf("%lld\n",quickpow(a,tb,c));}}
(4)如果a,b都是高精度,则:

对a,进行a%=c;处理,将a,变为int 型

对b,用上述公式处理,将b,变为int型

然后用快速幂,就可以了