HDU 5391 Zball in Tina Town(数论——威尔逊定理)——BestCoder Round #51(div.2)
Zball in Tina Town
Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
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.
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
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)
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附上该题对应的中文题
Zball in Tina Town
Time Limit: 3000/1500 MS (Java/Others)
Memory Limit: 262144/262144 K (Java/Others)
问题描述
Tina Town 是一个善良友好的地方,这里的每一个人都互相关心。 Tina有一个球，它的名字叫zball。zball很神奇，它会每天变大。在第一天的时候，它会变大1倍。在第二天的时候，它会变大2倍。在第n天的时候，它会变大n倍。 zball原来的体积是1。Tina想知道，zball在第n−1天时的体积对n取模是多大呢？ Tina是一个蠢蠢的女孩子，当然不会算啦，所以她请你帮她算出这个答案呢。
输入描述
第一行一个正整数T，表示数据组数 接下来T行，每行一个正整数n，意义如题面所述 T≤105,2≤n≤109
输出描述
对于每组数据，输出一个正整数，表示答案。
输入样例
2 3 10
输出样例
2 0
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出题人的解题思路：
1000 Zball in Tina Town
出题人:wuzhuangtai00
这题就是求 (n−1)! mod n
如果n为合数，显然答案为0.
如果n为素数，那么由威尔逊定理可得答案为 n−1
注意有个trick为 n = 4.
脑到用时方恨少，说实话，看着一位位大牛们把这题A出来，我就觉得自己为何如此蠢，虽然看懂了题目是让我们求(n-1)!对n取模的结果，但是着实不知道如何着手，甚至我还查到了阶乘的近似公式(stirling公式)
然而并没有任何用处。
看了出题人的解题报告之后，我才惊呆了，居然用了数论四大定理，我只能说我学艺不精，当初貌似有看到过威尔逊定理，但是被我默默忽略了，因为百度百科关于威尔逊定理有这么一句描述的话“由于阶乘是呈爆炸增长的，其结论对于实际操作意义不大”。好吧，我醉了，还是一起来温故一下这个定理吧
( p-1)! ≡ -1 ( mod p )
即(p-1)!+1能被p整除(当且仅当p为素数)
在此先不予证明，想要看证明的可以看一下百度百科(链接)，虽然表示我看得云里雾里
回到出题人的解题报告，素数的问题解决了，那我们看n为合数的情况，除了4以外，因为每一个合数都存在比它小的约数能组成该合数，所以(n-1)!必定是能够被n整除的，比如12，它能由3*4构成，所以11!必定能被12整除
而4，只有一个2，不够组成4，所以4要单独考虑。
关于上述判断素数的方法，普通的判断当然ok
for(int i=2;i*i<=n;i++)
if(n%i==0)
break;
而你用费马检测的方法也是行的
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
#define ll __int64
using namespace std;
const int N = 500005;
const int inf = 1000000000;
const int mod = 258280327;
//快速幂
ll Quick_Mod(ll a, ll b, ll mod)
{
ll res = 1,term = a % mod;
while(b)
{
if(b & 1) res = (res * term) % mod;
term = (term * term) % mod;
b >>= 1;
}
return res;
}
//费马检测
bool Is_Prime(ll n)
{
int i;
for(i = 0;i < 5;i++)//随机次数
if(Quick_Mod(1 + rand() % (n - 1),n - 1,n) != 1)
break;
if(i == 5) return true;
return false;
}
int main()
{
int t;
__int64 n;
scanf("%d",&t);
while(t--)
{
scanf("%I64d",&n);
if(n==4)
puts("2");
else if(Is_Prime(n))
printf("%I64d\n",n-1);
else
puts("0");
}
return 0;
}菜鸟成长记