HDOJ 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): 817 Accepted Submission(s): 471 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≤$$10^5$$,2≤n≤$$10^9$$ Output For each test case, output an integer representing the answer. Sample Input 2 3 10 Sample Output 2 Source BestCoder Round #51 (div.2) Recommend hujie | We have carefully selected several similar problems for you: 5395 5394 5393 5392 5390
本题关键在于理解题意，在BestCoder中文描述中是第n天变大n倍，这样不就是n+1倍了吗，当时一直按这个思路，以为是A176787这个序列，后来看英文以及别人的解释才明白。
On the n-th day,it will become n times as large as the size on the (n-1)-th day.所以第n天的大小就是第n-1天的n倍，也就是n!。 最终结果即为$$(n-1)!modn$$，如果n为合数，n一定能分解成n以内的某些个数相乘，所以结果为0；如果n为素数，根据威尔逊定理，有
所以结果为$$-1modn=n-1$$。
当n=2和4时，需要特殊处理。完整代码如下：
[cpp]
#include<iostream>
#include<cstdio>
using namespace std;
bool IsPrime(int x)
{
for (int i = 2; i*i <= x; i++)
if (x%i == 0)
return false;
return true;
}
int main()
{
int t, n;
scanf("%d", &t);
while (t–)
{
scanf("%d", &n);
if (n == 2)
printf("1n");
else if (n == 4)
printf("2n");
else if (IsPrime(n))
printf("%dn", n – 1);
else
printf("0n");
}
return 0;
}
[/cpp]
本代码提交AC，用时967MS，内存1576K。]]>