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.
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
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
题意:求 从1乘到n-1模n的结果。
只要判断n是否是素数,如果是就输出n-1,否则输出0。n=4,不满足这规律,要特判。
1 #include<cstdio> 2 #include<cstring> 3 #include<map> 4 #include<cmath> 5 using namespace std; 6 int zj[100000]; 7 int ssb[100000]; 8 int pn; 9 void p() 10 { 11 int i,j; 12 zj[1]=0; 13 zj[0]=0; 14 pn=0; 15 for (i=2;i<=100000;i++) 16 { 17 if (!zj[i]) ssb[pn++]=i; 18 for (j=0;j<pn;j++) 19 { 20 if (i*ssb[j]>100000) break; 21 zj[i*ssb[j]]=1; 22 if (i%ssb[j]==0) break; 23 } 24 } 25 } 26 int main() 27 { 28 p(); 29 int n,m,t,i,flag; 30 scanf("%d",&t); 31 while (t--) 32 { 33 flag=0; 34 scanf("%d",&n); 35 if (n==3||n==4) {printf("2\n");continue;} 36 if (n<=100000) 37 { 38 if (zj[n]) printf("0\n"); 39 else printf("%d\n",n-1); 40 continue; 41 } 42 flag=0; 43 for (i=0;i<pn;i++) 44 { 45 if (ssb[i]*ssb[i]>n||flag) break; 46 if (n%ssb[i]==0) flag=1; 47 } 48 if (flag) printf("0\n"); 49 else printf("%d\n",n-1); 50 } 51 }
11