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 become2 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

2310

 

Sample Output

20

 

Source

BestCoder Round #51 (div.2)

 

n是质数为N-1，不是为0，特判4

#include<iostream>#include<cstdio>#include<cstring>#include<vector>using namespace std;int check[100000];vector<int> 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;}