hdu5391Wilson定理
hdu5391威尔逊定理
Total Submission(s): 276 Accepted Submission(s): 168
开始一直超时,问题出在判断素数上。
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 becomes1 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.
Tina has a ball called zball. Zball is magic. It grows larger every day. On the first day, it becomes
Input
The first line of input contains an integer T ,
representing the number of cases.
The followingT lines,
each line contains an integer n ,
according to the description.
T≤105,2≤n≤109
The following
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
如果n为合数,显然答案为0.
如果n为素数,那么由威尔逊定理可得答案为 n−1
注意有个trick为 n = 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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