Zball in Tina Town
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
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.
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
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
2
3
10
Sample Output
2
0
Source
思路:当数为素数时,威尔逊;
当数为非素数,4输出2;
别的输出0;
#include<bits/stdc++.h> using namespace std; #define ll __int64 #define esp 0.00000000001 const int N=1e3+10,M=1e6+10,inf=1e9+10,mod=1000000007; int prime(int n) { if(n<=1) return 0; if(n==2) return 1; if(n%2==0) return 0; int k, upperBound=n/2; for(k=3; k<=upperBound; k+=2) { upperBound=n/k; if(n%k==0) return 0; } return 1; } int main() { int x,y,z,i,t; int T; scanf("%d",&T); while(T--) { scanf("%d",&x); if(x==4) printf("2 "); else if(prime(x)) printf("%d ",x-1); else printf("0 "); } return 0; }