hdu 5391 Zball in Tina Town

Zball in Tina Town

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 307    Accepted Submission(s): 187

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

 

Source

BestCoder Round #51 (div.2)

 

n是质数为N-1，不是为0，特判4

#include
#include
#include
#include
using namespace std;

int check[100000];
vector pri;
void init(){
    memset(check,0,sizeof(check));
    pri.clear();
    for(int i = 2;i < 100000; i++){
        if(check[i] == 0) {
            pri.push_back(i);
            for(int j = i+i;j < 100000; j += i){
                check[j] = 1;
            }
        }
    }
}

int main(){
    int n;
    int t;
    init();
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        if(n == 4){
            printf("2\n");
            continue;
        }
        if(n < 100000){
            if(check[n] == 0){
                printf("%d\n",n-1);
            }
            else printf("0\n");
        }
        else {
            int flag = 1;
            for(int i = 0;i < pri.size() && flag && pri[i]*pri[i] <= n ; i++){
                if(n % pri[i] == 0){
                    flag = 0;
                }
            }
            if(flag)printf("%d\n",n-1);
            else printf("0\n");
        }
    }
    return 0;
}