hdu 1395 2^x mod n = 1
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11835 Accepted Submission(s): 3684
Problem Description
Give a number n, find the minimum x(x>0) that satisfies 2^x mod n = 1.
Input
One positive integer on each line, the value of n.
Output
If the minimum x exists, print a line with 2^x mod n = 1.
Print 2^? mod n = 1 otherwise.
You should replace x and n with specific numbers.
Sample Input
2
5
Sample Output
2^? mod 2 = 1
2^4 mod 5 = 1
/*
方法一：暴力搜索 + 同余定理
*/
#include<cstdio>
int main()
{
int m,n,t;
while(scanf("%d",&n)!=EOF)
{
m=n%2?2:0;
t=1;
while(m!=1&&m)
{
m=m*2;
m%=n;
t++;
}
if(n%2==0||n==1) printf("2^? mod %d = 1\n",n);
else printf("2^%d mod %d = 1\n",t,n);
}
return 0;
}
/*
方法二： 蒙哥马利幂模运算
*/
#include<cstdio>
int Montgomery(long long a,int b,int c)
{
long long ans=1;
while(b)
{
if(b&1)
ans=(ans*a)%c;
b>>=1;
a=(a*a)%c;
}
return ans;
}
int main()
{
int m,i;
while(scanf("%d",&m)!=EOF)
{
if(m%2==0||m<=1) //如果不加 m<=1 会出现 TLE .
printf("2^? mod %d = 1\n",m);
else
for(i=1; ; i++)
{
if(Montgomery(2,i,m)==1)
{
printf("2^%d mod %d = 1\n",i,m);
break;
}
}
}
return 0;
}
