标签：数论 hdu 杭电 acm
威尔逊定理
在初等数论中，威尔逊定理给出了判定一个自然数是否为素数的充分必要条件。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.
T≤105,2≤n≤109
Output
For each test case, output an integer representing the answer.
Sample Input
Sample Output
Source
题解：
这题就是求 (n?1)! mod n(n-1)!modn(n?1)!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-1n?1
注意有个trick为 nnn =
4
代码：
#include<iostream>
#include<cmath>
#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<<ans<<endl;
}
return 0;
}
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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hdu5391威尔逊定理
标签：数论 hdu 杭电 acm
原文地址：http://blog.csdn.net/qq_27803491/article/details/47693147
