HDU 1395 2^x mod n = 1

2^x mod n = 1

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6543 Accepted Submission(s): 1961


[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

[align=left]Recommend[/align]
Ignatius.L
//hash的应用，判重复，

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;
bool h[10000];
int main()
{
int n,t,k;
while(scanf("%d",&n)!=EOF)
{
memset(h,0,n*sizeof(bool));
k=1;t=2;h[2]=1;
while(t%n!=1)
{
k++;
t=t<<1;
t=t%n;
if(h[t]) break;
h[t]=1;
}
if(t%n!=1)
printf("2^? mod %d = 1\n",n);
else
printf("2^%d mod %d = 1\n",k,n);
}
return 0;
}

// 加进去点东西，不过貌似没加快速度呀
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
using namespace std;
bool h[10000];
int main()
{
int n,t,k;
while(scanf("%d",&n)!=EOF)
{
if(n%2==0||n==1)//这里可以这样理解 n%2=0，n是偶数， 2^x是偶数，2^x%n也是偶数,所以不可 能 会 // 是1
{printf("2^? mod %d = 1\n",n);continue;}
memset(h,0,n*sizeof(bool));
k=1;t=2;h[2]=1;
while(t%n!=1)
{
k++;
t=t<<1;
t=t%n;
if(h[t]) break;
h[t]=1;
}
if(t%n!=1)
printf("2^? mod %d = 1\n",n);
else
printf("2^%d mod %d = 1\n",k,n);
}
return 0;
}