1600: Big Mod (大數冪模數)

Result	TIME Limit	MEMORY Limit	Run Times	AC Times	JUDGE
	3s	8192K	1756	336	Standard

Calculate

 

 

for large values of B, P, and M using an efficient algorithm. (That's right, this problem has a time dependency !!!.)

Input

Three integer values (in the order B, P, M) will be read one number per line. B and P are integers in the range 0 to 2147483647 inclusive. M is an integer in the range 1 to 46340 inclusive.

Output

The result of the computation. A single integer.

Sample Input

 

3181321717176532374859302938236123

Sample Output

 

13213195/*

*/

1. #include <stdio.h>
2. #include <memory>
3. int main ()
4. {
5. int a,b,n;
6. while (scanf("%d%d%d",&a,&b,&n)!=EOF)
7. {
8.   bool b2[35];
9.   memset (b2,0,sizeof(b2));
10.   int tmp=b,i,d=1;
11.   for ( i=0; tmp>0 ; i++)
12.   {
13.    b2[i]=tmp%2;
14.    tmp=tmp/2;
15.   }
16.   for (;i>=0;i--)
17.    {
18.    d=d*d%n;
19.    int aa=a%n;
20.    if ( b2[i]==1)
21.     d=d*aa%n;
22.    }
23.    printf("%d/n",d);
24. }
25. return 0;
26. }

方法二：超短代碼

#include <cstdio>
int b, p, m, k;
int main()
{
 while(scanf("%d%d%d", &b, &p, &m)!=EOF)
    {
  for(k=1,b=b%m; p ; p/=2,b=b*b%m)
   if(p%2) k=k*b%m;
  printf("%d\n", k%m);
 }
 return 0;
}

 

方法三：遞迴 時間稍慢

#include <cstdio>
int m;
int bigmod(int a,int b)
{
    if(!b)return 1;
    int d=bigmod(a,b>>1)%m;
    return  b&1?(d*d%m*a)%m:(d*d)%m;
}
int main ()
{
    int a,b;
    while (scanf("%d%d%d",&a,&b,&m)!=EOF)
    {
        int ans=bigmod(a%m,b)%m;
        printf("%d\n",ans);
    }
    return 0;
}

快速冪模數 最短代碼 for(k=1,a=a%m; b ; b>>=1,a=a*a%m) if(b&1) k=k*a%m; 求a^b(mod m) k%m是答案。