题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=1250

Hat's Fibonacci

Description

A Fibonacci sequence is calculated by adding the previous two members the sequence, with the first two members being both 1.
$F(1) = 1, F(2) = 1, F(3) = 1,F(4) = 1, F(n > 4) = F(n - 1) + F(n-2) + F(n-3) + F(n-4)$
Your task is to take a number as input, and print that Fibonacci number.

Input

Each line will contain an integers. Process to end of file.

Output

For each case, output the result in a line.

SampleInput

100

SampleOutput

4203968145672990846840663646

大数加法模板题。。

 #include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cassert>
#include<cstdio>
#include<vector>
#include<string>
#include<map>
#include<set>
using std::cin;
using std::max;
using std::cout;
using std::endl;
using std::string;
using std::istream;
using std::ostream;
#define sz(c) (int)(c).size()
#define all(c) (c).begin(), (c).end()
#define iter(c) decltype((c).begin())
#define cls(arr,val) memset(arr,val,sizeof(arr))
#define cpresent(c, e) (find(all(c), (e)) != (c).end())
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define fork(i, k, n) for (int i = (int)k; i <= (int)n; i++)
#define tr(c, i) for (iter(c) i = (c).begin(); i != (c).end(); ++i)
#define pb(e) push_back(e)
#define mp(a, b) make_pair(a, b)
struct BigN {
typedef unsigned long long ull;
static const int Max_N = ;
int len, data[Max_N];
BigN() { memset(data, , sizeof(data)), len = ; }
BigN(const int num) {
memset(data, , sizeof(data));
*this = num;
}
BigN(const char *num) {
memset(data, , sizeof(data));
*this = num;
}
void clear() { len = , memset(data, , sizeof(data)); }
BigN& clean(){ while (len > && !data[len - ]) len--; return *this; }
string str() const {
string res = "";
for (int i = len - ; ~i; i--) res += (char)(data[i] + '');
if (res == "") res = "";
res.reserve();
return res;
}
BigN operator = (const int num) {
int j = , i = num;
do data[j++] = i % ; while (i /= );
len = j;
return *this;
}
BigN operator = (const char *num) {
len = strlen(num);
for (int i = ; i < len; i++) data[i] = num[len - i - ] - '';
return *this;
}
BigN operator + (const BigN &x) const {
BigN res;
int n = max(len, x.len) + ;
for (int i = , g = ; i < n; i++) {
int c = data[i] + x.data[i] + g;
res.data[res.len++] = c % ;
g = c / ;
}
return res.clean();
}
BigN operator * (const BigN &x) const {
BigN res;
int n = x.len;
res.len = n + len;
for (int i = ; i < len; i++) {
for (int j = , g = ; j < n; j++) {
res.data[i + j] += data[i] * x.data[j];
}
}
for (int i = ; i < res.len - ; i++) {
res.data[i + ] += res.data[i] / ;
res.data[i] %= ;
}
return res.clean();
}
BigN operator * (const int num) const {
BigN res;
res.len = len + ;
for (int i = , g = ; i < len; i++) res.data[i] *= num;
for (int i = ; i < res.len - ; i++) {
res.data[i + ] += res.data[i] / ;
res.data[i] %= ;
}
return res.clean();
}
BigN operator - (const BigN &x) const {
assert(x <= *this);
BigN res;
for (int i = , g = ; i < len; i++) {
int c = data[i] - g;
if (i < x.len) c -= x.data[i];
if (c >= ) g = ;
else g = , c += ;
res.data[res.len++] = c;
}
return res.clean();
}
BigN operator / (const BigN &x) const {
BigN res, f = ;
for (int i = len - ; ~i; i--) {
f *= ;
f.data[] = data[i];
while (f >= x) {
f -= x;
res.data[i]++;
}
}
res.len = len;
return res.clean();
}
BigN operator % (const BigN &x) {
BigN res = *this / x;
res = *this - res * x;
return res;
}
BigN operator += (const BigN &x) { return *this = *this + x; }
BigN operator *= (const BigN &x) { return *this = *this * x; }
BigN operator -= (const BigN &x) { return *this = *this - x; }
BigN operator /= (const BigN &x) { return *this = *this / x; }
BigN operator %= (const BigN &x) { return *this = *this % x; }
bool operator < (const BigN &x) const {
if (len != x.len) return len < x.len;
for (int i = len - ; ~i; i--) {
if (data[i] != x.data[i]) return data[i] < x.data[i];
}
return false;
}
bool operator >(const BigN &x) const { return x < *this; }
bool operator<=(const BigN &x) const { return !(x < *this); }
bool operator>=(const BigN &x) const { return !(*this < x); }
bool operator!=(const BigN &x) const { return x < *this || *this < x; }
bool operator==(const BigN &x) const { return !(x < *this) && !(x > *this); }
friend istream& operator >> (istream &in, BigN &x) {
string src;
in >> src;
x = src.c_str();
return in;
}
friend ostream& operator << (ostream &out, const BigN &x) {
out << x.str();
return out;
}
}A[];
void solve(int n) {
fork(i, , ) A[i] = ;
if (n < ) cout << A[n] << endl;
else {
int x = ;
fork(i, , n) {
A[] = A[] + A[] + A[] + A[];
A[x] = A[];
if (++x == ) x = ;
}
cout << A[] << endl;
}
rep(i, ) A[i].clear();
}
int main() {
#ifdef LOCAL
freopen("in.txt", "r", stdin);
freopen("out.txt", "w+", stdout);
#endif
std::ios::sync_with_stdio(false);
int n;
while (cin >> n) {
solve(n);
}
return ;
}

hdu 1250 Hat's Fibonacci的更多相关文章

  1. HDU 1250 Hat's Fibonacci(大数相加)

    传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1250 Hat's Fibonacci Time Limit: 2000/1000 MS (Java/Ot ...

  2. HDU 1250 Hat's Fibonacci (递推、大数加法、string)

    Hat's Fibonacci Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)T ...

  3. hdu 1250 Hat's Fibonacci(高精度数)

    //  继续大数,哎.. Problem Description A Fibonacci sequence is calculated by adding the previous two membe ...

  4. HDOJ/HDU 1250 Hat's Fibonacci(大数~斐波拉契)

    Problem Description A Fibonacci sequence is calculated by adding the previous two members the sequen ...

  5. HDU 1250 Hat's Fibonacci(高精度)

    Problem Description A Fibonacci sequence is calculated by adding the previous two members the sequen ...

  6. hdu 1250 Hat's Fibonacci(java,简单,大数)

    题目 java做大数的题,真的是神器,来一道,秒一道~~~ import java.io.*; import java.util.*; import java.math.*; public class ...

  7. hdu 1250 Hat&#39;s Fibonacci

    pid=1250">点击此处就可以传送hdu 1250 Problem Description A Fibonacci sequence is calculated by adding ...

  8. Hat's Fibonacci(大数加法+直接暴力)

    题目连接:http://acm.hdu.edu.cn/showproblem.php?pid=1250 hdu1250: Hat's Fibonacci Time Limit: 2000/1000 M ...

  9. HDUOJ----1250 Hat's Fibonacci

    Hat's Fibonacci Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)T ...

随机推荐

  1. ionic cordova social media sharing plugin

    https://github.com/EddyVerbruggen/SocialSharing-PhoneGap-Plugin.git 从上面的连接下载 插件 1.肯定是要加入 下面的那个文件的吖 从 ...

  2. python线程使用场景 多线程下载

    http://blog.xiayf.cn/2015/09/11/parallelism-in-one-line http://python.jobbole.com/84327/ http://www. ...

  3. css选择器分类

    css选择器大致可以分为10大类: 1.元素选择器 如:p{} 2.类选择器 如:.xx{} 3.ID选择器 如:#xx{} 4.关联选择器 如:p a{} 5.子元素选择器 如:p>a{} 6 ...

  4. hbase日常操作及维护

    一,基本命令: 建表:create 'testtable','coulmn1','coulmn2' 也可以建表时加coulmn的属性如:create 'testtable',{NAME => ' ...

  5. poj2031 Building a Space Station

    这题目,用G++ WA,用C++ AC. 题目要求,现给出n个球,然后要使每两个球直接或者间接连通,可以在任意两球之间做管道(在表面),最后的要求是,如果使得都连通的话,管道最小长度是多少. 思路简单 ...

  6. 【风马一族_php】PHP与Mysql建立连接

    让php发出 Hi作为基础 http://www.cnblogs.com/sows/p/5990157.html 配置apache      ../apache/conf/httpd.conf 创建p ...

  7. .NET Web开发总结(三)

    第五章 ASP.NET 页面语法 本章详细讲解.NET页面的语法结构     一般情况下  一个ASP.NET页面要包括页面编译指令 HTML页面框架及Web窗体     服务器端控件  服务器端代码 ...

  8. css3内处理

    1.插入文字:content属性:          p:after{content:"内容"}          p:before{content:"内容"} ...

  9. ASP.Net MVC 5 高级编程 第7章 成员资格、授权和安全性

    第7章 成员资格.授权和安全性 7.1 安全性 ASP.NET MVC 提供了许多内置的保护机制(默认利用 HTML 辅助方法和Razor 语法进行 HTML编码以及请求验证等功能特性,以及通过基架构 ...

  10. JqueryMobile动态生成listView并实现刷新的两种方法

    本篇文章主要是对JqueryMobile动态生成listView并实现刷新的两种方法进行了介绍,需要的朋友可以过来参考下,希望对大家有所帮助 JqueryMobile动态生成listView并实现刷新 ...