hdu5391

Zball in Tina Town

Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1930 Accepted Submission(s): 977

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
2
3
10

Sample Output
2
0

昨晚比赛的这道题，题意就是求n-1的阶乘对n取余的值，因为n的最大值为10的9次方，当时感觉用阶乘会超时，于是就开始看有没有规律可寻，找了前21个数，把他们的结果都列出来了，但硬是没看出来规律，太笨了，后来比赛结束后学长说了一下规律，我天哪，我的数据结果完全符合【流汗】但就是没有往那方面想，看来还是做题少。废话不说了，试着找一下前21个数的结果，你会发现这个规律的。。。（该猜题的时候就要大胆猜）。

#include<stdio.h>#include<algorithm>#include<stdlib.h>#include<iostream>#include<string.h>#include<vector>#include<queue>#include<math.h>using namespace std;int _isprime(int m){    for(int i=2; i<=sqrt(m); i++)        if(m%i==0)            return 0;    return 1;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        int n;        int sum;        scanf("%d",&n);        if(n==1)            printf("0\n");        else  if(_isprime(n))            printf("%d\n",n-1);        else        {            if(n==4)                printf("2\n");            else                printf("0\n");        }    }    return 0;}