#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
#include<algorithm>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<stack>
//#include<bits/std c++.h>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
const LL MOD = 1e7 + 7;
const LL maxn = 1e6 + 131;
const LL Bits = 1e10;
const double Pi = acos(-1.0);
struct BigInt
{
vector<LL> val;
void Set(LL n) { val.clear(); val.push_back(n); }
void operator += (const BigInt a)
{
int i;
for(i = 0; i < val.size() && i < a.val.size() ; ++i)
val[i] = val[i] + a.val[i];
if(i < a.val.size()) for(; i < a.val.size(); ++i) val.push_back(a.val[i]);
int len = val.size();
for(i = 0; i < len; ++i)
{
if(val[i] >= Bits)
{
if(i == len-1) val.push_back(val[i]/Bits), val[i] %= Bits;
else {
val[i+1] += (val[i] / Bits);
val[i] %= Bits;
}
}
}
}
void operator *= (const LL k)
{
int i;
for(i = 0; i < val.size(); ++i) val[i] *= k;
i = 0;
while(i != val.size())
{
if(val[i] >= Bits)
{
if(i == val.size() -1)
{
val.push_back(val[i] / Bits);
val[i] %= Bits;
}
else
{
val[i+1] += (val[i] / Bits);
val[i] %= Bits;
}
}
++i;
}
}
void operator /= (const LL k)
{
for(int i = val.size() - 1; i >= 1; --i)
{
val[i-1] += (val[i] % k)*Bits;
val[i] /= k;
}
val[0] /= k;
int i = val.size() - 1;
while(val.size() > 1)
{
if(val[i] == 0) val.erase(--val.end());
else break;
i--;
}
}
void Show()
{
cout << val[val.size() - 1];
for(int i = val.size() - 2; i >= 0 ; --i)
{
LL k = Bits / 10;
while( k > 1 )
{
if(val[i] < k) cout << "0";
k /= 10;
}
cout << val[i];
}
cout << endl;
}
}; int main()
{
BigInt A, B;
LL n;
/*A.Set(1), B.Set(Bits - 1);
A += B;
A.Show();*/
//cout << Pi << endl;
while(cin >> n){
if(n < 4) cout << "1\n";
else {
n -= 4;
A.Set(1), B.Set(1);
LL i = 1, j = (n + Pi), Last;
while(i <= (LL)((n + Pi)/ Pi))
{
Last = j;
j = (n + Pi - i*Pi);
for(LL k = Last - 1; k >= j + 1; k--)
{
B *= (k+1);
B /= (i + k); }
B *= (j + 1);
B /= i;
i ++;
A += B;
}
A.Show();
}
}
return 0;
}

CodeVs 3150 (大数 + 递推)的更多相关文章

  1. HDU-1041-Computer Transformation,大数递推,水过~~

                                                                                  Computer Transformatio ...

  2. POJ 1737 Connected Graph (大数+递推)

    题目链接: http://poj.org/problem?id=1737 题意: 求 \(n\) 个点的无向简单(无重边无自环)连通图的个数.\((n<=50)\) 题解: 这题你甚至能OEIS ...

  3. UVa 10328 Coin Toss(Java大数+递推)

    https://vjudge.net/problem/UVA-10328 题意: 有H和T两个字符,现在要排成n位的字符串,求至少有k个字符连续的方案数. 思路:这道题目和ZOJ3747是差不多的,具 ...

  4. 1sting 大数 递推

    You will be given a string which only contains ‘1’; You can merge two adjacent ‘1’ to be ‘2’, or lea ...

  5. hdu 1133 Buy the Ticket (大数+递推)

    Buy the Ticket Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)To ...

  6. Tiling(递推+大数)

    Description In how many ways can you tile a 2xn rectangle by 2x1 or 2x2 tiles? Here is a sample tili ...

  7. Children’s Queue HDU 1297 递推+大数

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1297 题目大意: 有n个同学, 站成一排, 要求 女生最少是两个站在一起, 问有多少种排列方式. 题 ...

  8. 递推,大数存储E - Order Count

    Description If we connect 3 numbers with "<" and "=", there are 13 cases: 1) ...

  9. HOJ 2148&POJ 2680(DP递推,加大数运算)

    Computer Transformation Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 4561 Accepted: 17 ...

随机推荐

  1. Reduce:规约;Collector:收集、判断性终止函数、组函数、分组、分区

    Stream 的终止操作  一.规约 * reduce(T iden, BinaryOperator b) 可以将流中元素反复结合起来,得到一个值. 返回 T * reduce(BinaryOpera ...

  2. T-SQL 编程技巧

    Ø  T-SQL 编程是大多数程序员都会接触的,也是数据库编程必须掌握的技术.下面,是本人在工作或学习中积累的一些心得和技巧.主要包含以下内容: 1.   waitfor延时执行 2.   NOT 关 ...

  3. sqlserver二进制存储

    CREATE TABLE myTable_yq(Document varbinary(max),yq varchar(20)) --SELECT @xmlFileName = 'c:\TestXml. ...

  4. Eclipse下生成/编辑Java类图或时序图(UML)[转载]

    一 引用文章 1.[eclipse下生成Java类图和时序图,生成UML图(更完整版)](https://blog.csdn.net/guomainet309/article/details/5302 ...

  5. [C++]PAT乙级1002.写出这个数(20/20)

    /* 1002. 写出这个数 (20) 读入一个自然数n,计算其各位数字之和,用汉语拼音写出和的每一位数字. 输入格式:每个测试输入包含1个测试用例,即给出自然数n的值.这里保证n小于10^100. ...

  6. mui的switch开关的应用

    HTML: <!--mui的switch开关--> <div class="mui-content-padded"> <h5>switch开关m ...

  7. vue请求java服务端并返回数据

    最近在自学vue怎么与java进行数据交互.其实axios还是挺简单的,与ajax请求几乎一样,无外乎也就是要解决下跨域的问题. 废话不多说了,直接贴代码,一看就懂! //向springmvc Con ...

  8. HttpClient和HttpURLConnection的使用和区别

    https://www.cnblogs.com/liushuibufu/p/4140913.html 功能用法对比 从功能上对比,HttpURLConnection比HttpClient库要丰富很多, ...

  9. linux系统 户和账号操作

    1,基本操作要求 实现用户账号的管理,要完成的工作主要有如下几个方面: ·       用户账号的添加.删除与修改.·       用户口令的管理.·       用户组的管理. 2,用户账户添加删除 ...

  10. 安装mysql8.0.12以及修改密码和Navicat的连接

    mysql8.0+与安装其他版本不同一.安装mysql8.0.121.到官网https://www.mysql.com/  下载mysql-8.0.12-winx64.zip(不要.mis),直接解压 ...