C++博客-ACM___________________________-随笔分类-ACM ( 大数 )

HDU 1250 HDOJ 1250 Hat's Fibonacci ACM 1250 IN HDU

MiYu — Fri, 22 Oct 2010 12:41:00 GMT

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋

题目地址 :

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

题目描述:

Hat's FibonacciTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2208    Accepted Submission(s): 684


Problem 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.
 

Sample Input
100
 

Sample Output
4203968145672990846840663646


Note:
No generated Fibonacci number in excess of 2005 digits will be in the test data, ie. F(20) = 66526 has 5 digits.
 

 
 /*
Mail to   : miyubai@gamil.com
MyBlog    : baiyun.me
Link      : http://www.cnblogs.com/MiYu  || http://www.cppblog.com/MiYu
Author By : MiYu
Test      : 1
Complier  : g++ mingw32-3.4.2
Program   : Hat's Fibonacci
Doc Name  : HDU_1250
*/


/*

高精度的水题 , JAVA  水过, 当然 用模板也一样 

*/

package main;
import java.util.*;
import java.math.*;
import java.io.*;
public class Main {
    public static void main(String[] args) {
        Scanner cin = new Scanner ( System.in );
        while ( cin.hasNext() ) {
            int N = cin.nextInt();
            if ( N <= 4 ) {
                System.out.println ( 1 );
            } else {
                BigInteger one = new BigInteger ( "1" );
                BigInteger two = new BigInteger ( "1" );
                BigInteger three = new BigInteger ( "1" );
                BigInteger four = new BigInteger ( "1" );
                BigInteger res = new BigInteger ( "1" );
                BigInteger temp = new BigInteger ( "1" );
                for ( int i = 4; i < N; ++ i ) {
                    res = one.add(two).add(three).add(four);
                    temp = res;
                    one= two;
                    two = three;
                    three = four;
                    four = temp;
                }
                System.out.println( res );
            }
        }
    }

}

MiYu 2010-10-22 20:41 发表评论

HDOJ 1063 HDU 1063 Exponentiation JAVA 编写 ACM 1063 IN HDU

MiYu — Fri, 10 Sep 2010 09:01:00 GMT

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋

题目地址: http://acm.hdu.edu.cn/showproblem.php?pid=1063

题目描述:

Exponentiation

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1696 Accepted Submission(s): 376

Problem Description

Problems involving the computation of exact values of very large magnitude and precision are common. For example, the computation of the national debt is a taxing experience for many computer systems.

This problem requires that you write a program to compute the exact value of Rⁿ where R is a real number ( 0.0 < R < 99.999 ) and n is an integer such that 0 < n <= 25.

Input

The input will consist of a set of pairs of values for R and n. The R value will occupy columns 1 through 6, and the n value will be in columns 8 and 9.

Output

The output will consist of one line for each line of input giving the exact value of R^n. Leading zeros should be suppressed in the output. Insignificant trailing zeros must not be printed. Don't print the decimal point if the result is an integer.

Sample Input

95.123 12 0.4321 20 5.1234 15 6.7592  9 98.999 10 1.0100 12

Sample Output

548815620517731830194541.899025343415715973535967221869852721 .00000005148554641076956121994511276767154838481760200726351203835429763013462401 43992025569.928573701266488041146654993318703707511666295476720493953024 29448126.764121021618164430206909037173276672 90429072743629540498.107596019456651774561044010001 1.126825030131969720661201

这两天突然想学习下JAVA 的大数就学习了一下, 这道题在PKU A掉了的, 不过在 HDU 一直没过, 不知道有什么恶心的数据把我给恶心掉了, 很郁闷.

不过用JAVA直接 1 A 了........... 很简单的说

代码如下 :

/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
import java.util.Scanner;
import java.math.*;

/**
*
* @author baiyun
*/
public class Main {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        Scanner in = new Scanner ( System.in );
        BigDecimal num;
        int exp;
        while ( in.hasNext() ){
            num = in.nextBigDecimal();
            exp = in.nextInt();
            num = num.pow ( exp );
            num = num.stripTrailingZeros();
            String out = num.toPlainString();
            if ( out.charAt(0) == '0' )
                 out = out.substring(1);
            System.out.println( out );
        }
    }
}

MiYu 2010-09-10 17:01 发表评论

HDOJ HDU 1715 大菲波数 ACM 1715 IN HDU

MiYu — Sun, 08 Aug 2010 14:38:00 GMT

MiYu原创, 转帖请注明 : 转载自 ______________白白の屋

题目地址:
http://acm.hdu.edu.cn/showproblem.php?pid=1715
题目描述:

Problem Description
Fibonacci数列，定义如下：
f(1)=f(2)=1
f(n)=f(n-1)+f(n-2) n>=3。
计算第n项Fibonacci数值。

Input
输入第一行为一个整数N，接下来N行为整数Pi（1<=Pi<=1000）。

Output
输出为N行，每行为对应的f(Pi)。

Sample Input
5
1
2
3
4
5

Sample Output
1
1
2
3
5

看题目就知道是大数了..........直接模板吧 .没什么好说的.
代码如下:

#include <iostream>
#include <string>
using namespace std;

string Add(string fNum,string sNum)
{
        if( fNum.length() < sNum.length() ) fNum.swap(sNum);
        string A = "0";
        A += fNum;
        for ( int i=1;i<=fNum.length();i++)
                if(i<=sNum.length())
                        A[A.length()-i]+=sNum[sNum.length()-i]-'0';
        for( int i=1;i<A.length();i++)
        {
                if(A[A.length()-i]>'9')
                { A[A.length()-i]-=10;A[A.length()-i-1]+=1; }
        }
        while(A[0]=='0') A.erase(0,1);
        return A;
}

string F[1001] = { "0", "1", "1" };
void setNum ()
{
     for ( int i = 3; i != 1001; ++ i )
     {
           F[i] = Add ( F[i-1] , F[i-2] ) ;
     }
}
int main ()
{
    int T;
    setNum ();
    while ( cin >> T )
    {
            while ( T -- )
            {
                   int N;
                   cin >> N;
                   cout << F[N] << endl;
            }
    }
    return 0;
}

MiYu 2010-08-08 22:38 发表评论