hdu5391威尔逊定理

威尔逊定理

初等数论中,威尔逊定理给出了判定一个自然数是否为素数充分必要条件。即:当且仅当p为素数时:( p -1 )! ≡ -1 ( mod p ),但是由于阶乘是呈爆炸增长的,其结论对于实际操作意义不大。

hdu5391用到了这一数论定理。

Zball in Tina Town

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 276    Accepted Submission(s): 168


Problem Description
Tina Town is a friendly place. People there care about each other.

Tina has a ball called zball. Zball is magic. It grows larger every day. On the first day, it becomes  1 time as large as its original size. On the second day,it will become  2 times as large as the size on the first day. On the n-th day,it will become  n times as large as the size on the (n-1)-th day. Tina want to know its size on the (n-1)-th day modulo n.
 

Input
The first line of input contains an integer  T, representing the number of cases.

The following  T lines, each line contains an integer  n, according to the description.
T105,2n109
 

Output
For each test case, output an integer representing the answer.
 

Sample Input
 
   
2 3 10
 

Sample Output
 
   
2 0
 

Source
BestCoder Round #51 (div.2)
 
题解:

这题就是求 (n−1)!  mod  n(n-1)!modn(n1)!mod n

如果nnn为合数,显然答案为0.

如果nnn为素数,那么由威尔逊定理可得答案为 n−1n-1n1

注意有个trick为 nnn = 4

代码:
#include
#include
#define ll long long 

using namespace std;

int t;
int n,ans;

bool Isprime(int m)
{
    int i,isqrt=(int)sqrt(m);
    bool pd=true;
    if(m==2)pd=true;
    else if(m%2==0)pd=false;
    else 
    for(i=3;i<=isqrt;i+=2){
        if(m%i==0){
            pd=false;
            break;
        }
    }
    return pd;
}

int main()
{
    cin>>t;
    while(t--){
        cin>>n;
        if(n==4)ans=2;
        else if(Isprime(n))ans=n-1;
        else ans=0;
        cout<

开始一直超时,问题出在判断素数上。
一定要注意素数判定问题。

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