HDOJ 题目1395 2^x mod n = 1（水题 易错）

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 11712    Accepted Submission(s): 3649


[align=left]Problem Description[/align]
Give a number n, find the minimum x(x>0) that satisfies 2^x mod n = 1.

 

[align=left]Input[/align]
One positive integer on each line, the value of n.

 

[align=left]Output[/align]
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.

 

[align=left]Sample Input[/align]

2
5

 

[align=left]Sample Output[/align]

2^? mod 2 = 1
2^4 mod 5 = 1

 

[align=left]Author[/align]
MA, Xiao
 

[align=left]Source[/align]
ZOJ Monthly, February 2003

 
ac代码

#include<stdio.h>
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
if(n%2==0||n==1)//注意没有满足的情况下的条件，n==1，不要忘了
{
printf("2^? mod %d = 1\n",n);
}
else
{
int s=1,d=0;
while(1)
{
s*=2;
d++;
s%=n;//注意这儿，应用剩余定理，去掉会超时
if(s%n==1)
{
printf("2^%d mod %d = 1\n",d,n);
break;
}
}
}
}
}