明显是无良心的数学round= =

1000 Zball in Tina Town

 #include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<stack>
#include<queue>
#include<cstring>
#define PAU putchar(' ')
#define ENT putchar('\n')
#define Safe 6
using namespace std;
typedef long long LL;
const int maxn=+;
LL gcd(LL a, LL b){
return b == ? a : gcd(b, a % b);
}
LL n, x, mx;
LL mul(LL a, LL b, LL p){
LL tmp = (a * b - (LL)((long double)a / p * b + 1e-) * p);
return tmp < ? tmp + p : tmp;
}
LL pow(LL a, LL b, LL p){
LL ans = ; a %= p;
for(LL i = b; i; i >>= , a = mul(a, a, p))
if(i & ) ans = mul(ans, a, p);
return ans;
}
bool check(LL a, LL n, LL r, LL s){
LL ans = pow(a, r, n), p = ans;
for(int i = ; i <= s; i ++){
ans = mul(ans, ans, n);
if(ans == && p != && p != n - ) return true;
p = ans;
}
if(ans != )return true;
return false;
}
bool MR(LL n){
if(n <= ) return false;
if(n == ) return true;
if(n % == ) return false;
LL r = n - , s = ;
while(r % == ) r /= , s ++;
for(int i = ; i < Safe; i ++)
if(check(rand() % (n - ) + , n, r, s)) return false;
return true;
}
inline LL read(){
LL x=;bool sig=;char ch=getchar();
for(;!isdigit(ch);ch=getchar())if(ch=='-')sig=;
for(;isdigit(ch);ch=getchar())x=*x+ch-'';
return sig?x:-x;
}
inline void write(LL x){
if(x==){putchar('');return;}if(x<)putchar('-'),x=-x;
int len=;static LL buf[];while(x)buf[len++]=x%,x/=;
for(int i=len-;i>=;i--)putchar(buf[i]+'');return;
}
int T;
int main(){
T=read();LL x;
while(T--){
x=read();
if(x==){write();ENT;continue;}
if(x==){write();ENT;continue;}
if(MR(x))write(x-),ENT;
else write(),ENT;
}
// for(LL x=1;x<=100;x++){
// LL fac=1;
// for(int i=2;i<x;i++)fac=fac*i%x;write(x);PAU;write(fac);ENT;
// }
return ;
}

然后这道题全场都被至少扣了一分,不明觉厉= =

1001 Infoplane in Tina Town

显然不可写= =div1都没有人过= =

BestCoder Round #51 (div.2)的更多相关文章

  1. 判素数+找规律 BestCoder Round #51 (div.2) 1001 Zball in Tina Town

    题目传送门 /* 题意: 求(n-1)! mod n 数论:没啥意思,打个表能发现规律,但坑点是4时要特判! */ /***************************************** ...

  2. hdu 5636 搜索 BestCoder Round #74 (div.2)

    Shortest Path  Accepts: 40  Submissions: 610  Time Limit: 4000/2000 MS (Java/Others)  Memory Limit: ...

  3. BestCoder Round #69 (div.2) Baby Ming and Weight lifting(hdu 5610)

    Baby Ming and Weight lifting Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K ( ...

  4. BestCoder Round #68 (div.2) tree(hdu 5606)

    tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submis ...

  5. BestCoder Round #11 (Div. 2) 题解

    HDOJ5054 Alice and Bob Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/O ...

  6. hdu5635 BestCoder Round #74 (div.2)

    LCP Array  Accepts: 131  Submissions: 1352  Time Limit: 4000/2000 MS (Java/Others)  Memory Limit: 13 ...

  7. hdu5634 BestCoder Round #73 (div.1)

    Rikka with Phi  Accepts: 5  Submissions: 66  Time Limit: 16000/8000 MS (Java/Others)  Memory Limit: ...

  8. hdu5631 BestCoder Round #73 (div.2)

    Rikka with Graph  Accepts: 123  Submissions: 525  Time Limit: 2000/1000 MS (Java/Others)  Memory Lim ...

  9. hdu5630 BestCoder Round #73 (div.2)

    Rikka with Chess  Accepts: 393  Submissions: 548  Time Limit: 2000/1000 MS (Java/Others)  Memory Lim ...

随机推荐

  1. iOS UIScrollView 你可能不知道的奇技淫巧

    iOS 的 UIScrollView 可以说是十分强大,巧妙地运用它可以得到一些意想不到的效果.本文将举几个 ScrollView 不常见运用的例子. 自带信息应用 这个界面既可以上下卷动,也可以左右 ...

  2. Oracler读取各种格式的相关日期格式

    CREATE OR REPLACE Package Pkg_Stm_Date As     --Purpose:相关日期处理功能包          --获取某一天是第几周     Function ...

  3. jdk配置及maven配置

    jdk配置及maven配置 >>>>>>>>>>>>>>>>>>>>>&g ...

  4. matlab切比雪夫拟合

    matlab中没有切比雪夫拟合的现成算法,这里把我程序中的这部分抽出来,说一下. 1.首先是切比雪夫计算式 function [ res ] = ChebyShev(num,i) res=; else ...

  5. 10.26 noip模拟试题

    enc[问题背景]zhx 和他的妹子聊天.[问题描述]考虑一种简单的加密算法.假定所有句子都由小写英文字母构成,对于每一个字母,我们将它唯一地映射到另一个字母.例如考虑映射规则:a->b, b- ...

  6. 重新看php数组

    闲来有空,最近看php手册数组这块,对于array_values() 还是第一次接触,array_values是不保留键名,只有键值的函数,还有一个作用就是  重新索引. unset() 函数,是删除 ...

  7. List转xml

    1. List<Model> list = new List<Model>(); Model zj = new Model(); zj.id = ; zj.name = &qu ...

  8. WPF RichTextBox 如何滚动到光标所在位置、滚动条操作

    1.获取当前滚动条位置 //获取当前滚动条位置 richTextBox.VerticalOffset; richTextBox.HorizontalOffset; //获取当前光标位置 richTex ...

  9. java中json转xml

    参考:http://heipark.iteye.com/blog/1394844 需要json-lib-2.1-jdk15.jar和xom-1.2.5.jar,maven pom.xml如下: xml ...

  10. 某PHP代码加密

    <?php /* 本程序已加密: 2014-11-15 10:10:11 */ xs_run('JGxosS9QplmqLA6qjYo/LiX5ecUe0DH7p42Ww/Mdkf5/ybZDs ...