HDU1250

Java大数相加（多个大数相加）-hdu1250

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1250题目描述：题目大意是：已知一个Hat'sFibonacci序列，该序列满足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);此题用java的BigInteger处理很方便，但是因为我在处理F(n>4)=F(n-1)+F(n
weixin_30758821·2020-09-13 13:02

【Java / python】高精度数运算（大数类）训练

（大数阶乘）【HDU1047】IntegerInquiry（大数加法）【HDU1063】Exponentiation（实数高精度幂）【HDU1133】BuytheTicket（卡特兰数）【HDU1250
有所为，无所畏·2020-08-11 10:30

hdu1250 Hat's Fibonacci

Hat'sFibonacciTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):12633AcceptedSubmission(s):4241ProblemDescriptionAFibonaccisequenceiscalculatedbyaddingtheprevio
popcjz·2020-07-04 12:02

hdu1250

转自：http://blog.csdn.net/ultimater/article/details/7916412这个大家都懂，是个水题，知道用字符串作加法的这个肯定能懂，从左到右一位一位地加，题目中说答案不会超过2005个数字，而我一个int存了8位，所以可以确定数组的第二维最多开个255就行，而1维嘛，10的2006次方大约等于2的7000多次方，所以开个8000足够；LANGUAGE:C++
二手的玫瑰·2020-07-04 12:13

HDU1250 Hat's Fibonacci、1042 N! 、1002 A + B Problem II(大数练习)

A-A+BProblemIITimeLimit:1000MSMemoryLimit:32768KB64bitIOFormat:%I64d&%I64uSubmitStatusDescriptionIhaveaverysimpleproblemforyou.GiventwointegersAandB,yourjobistocalculatetheSumofA+B.InputThefirstlineof
大白QQly成长日记·2018-07-16 11:56

hdu1250 Hat's Fibonacci 斐波那契数列与大数加法

Hat'sFibonacciTimeLimit:2000/1000MS(Java/Others)MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):12254AcceptedSubmission(s):4105ProblemDescriptionAFibonaccisequenceiscalculatedbyaddingtheprevio
Dave_L·2018-01-03 09:16

HDU1250大数+斐波那契数列

DescriptionAFibonaccisequenceiscalculatedbyaddingtheprevioustwomembersthesequence,withthefirsttwomembersbeingboth1.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)Yourtaskistotakeanumber
vector_M·2016-02-18 16:00

HDU1250：Hat's Fibonacci

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) =
·2015-11-12 20:56

hdu 1250(java大数)

//java大数import java.util.Scanner;import java.math.BigInteger;public class hdu1250{ public static void
·2015-11-11 05:28

斐波那契数列hdu1250

Hat'sFibonacciCrawlinginprocess...CrawlingfailedTimeLimit:1000MS    MemoryLimit:32768KB    64bitIOFormat:%I64d&%I64u       DescriptionAFibonaccisequenceiscalculatedbyaddingtheprevioustwomembersthesequ
zcmartin2014214283·2015-08-20 22:00

hdu1250 Hat's Fibonacci 高精度

题目(http://acm.hdu.edu.cn/showproblem.php?pid=1250)ProblemDescription AFibonaccisequenceiscalculatedbyaddingtheprevioustwomembersthesequence,withthefirsttwomembersbeingboth1. F(1)=1,F(2)=1,F(3)=1,F(4)=
aonaigayiximasi·2015-08-13 17:00

hdu1250

链接：点击打开链接题意：大数斐波那契代码：#include #include #include usingnamespacestd; #defineN10000 #defineM300 #definemod100000000 inta[N][M]; voidbignumber(){ inti,j,t; a[1][1]=a[2][1]=a[3][1]=a[4][1]=1; for(i=5;i=1){
stay_accept·2015-08-13 13:00

HDU1250 Hat's Fibonacci 【大数】

Hat'sFibonacciTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):7616    AcceptedSubmission(s):2474ProblemDescriptionAFibonaccisequenceiscalculatedbyaddingth
u012846486·2014-08-28 17:00

HDU1250 Hat's Fibonacci 大数运算

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1250分析：大数运算，模拟加法即可。实现代码如下：#include #include usingnamespacestd; intf[6][2501]; intmain() { intn,i,j,k; while(scanf("%d",&n)!=-1) { if(n==0) { printf("0\n"
AC_Gibson·2014-08-07 17:00

HDU1250：Hat's Fibonacci

ProblemDescriptionAFibonaccisequenceiscalculatedbyaddingtheprevioustwomembersthesequence,withthefirsttwomembersbeingboth1.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)Yourtaskistotake
libin56842·2013-07-22 14:00

hdu1250 Hat's Fibonacci(高精度加)

Hat'sFibonacciTimeLimit:2000/1000MS(Java/Others)    MemoryLimit:65536/32768K(Java/Others)TotalSubmission(s):5097    AcceptedSubmission(s):1713ProblemDescriptionAFibonaccisequenceiscalculatedbyaddingth
yew1eb·2013-05-24 22:00

nyoj-114

大数斐波那契数列跟hdu1250相似，不够更难做一点#include #include #defineN103 #defineM800 #definebase100000000//进制 inta[N][
hong0220·2013-04-07 17:00

HDU1250 大整数问题

/*按照大整数的模版来；这个类型的斐波纳数列是前四项和，那么我们就讲其称为s1,s2,s3,s4，那么四项和即为s1+s2与s3+s4的和；运用一个知识点，就是二维数组str[n][m]的第i行的起始地址为str[i]；代码如下：*/#include #include intF[7061][550],c[550];//7060时位数为2012 constintBase=10000; //万进制in
i_fuqiang·2013-01-10 21:00

hdu1250

#includeusingnamespacestd;intmain(){inta[5][2010];inti,j;inttemp;intn,len,k;while(cin>>n){memset(a,0,sizeof(a));a[0][0]=a[1][0]=a[2][0]=a[3][0]=1;len=1;for(i=4;i=0;len--)cout<
fangzhiyang·2012-03-04 20:00

hdu 1250

hdu1250/* * 1250.cpp *   *  Created on: 2010-10-3 *      Author: wyiu */#include #include using namespace
wyiu·2010-10-03 17:00
上一页 1 下一页
按字母分类： ABCDEFGHIJKLMNOPQRSTUVWXYZ其他