Fighting_Dream

/*hdu 1999 不可摸数Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8274    Accepted Submission(s): 2122Problem Descriptions(n)是正整数n的真因子之和，

A sequence of numbersTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4041    Accepted Submission(s): 1240Problem DescriptionXinlv wr

今天突然看到了OJ上的大数类题目，由于学习了一点大数类的知识。果断水了6道题。。。。。。都是非常基础的。就当的练手的吧。学到的只是一些大数类的基本操作。以后多做点这样的题，争取熟练运用水大数题。。。大数阶乘题目链接：http://acm.nyist.net/JudgeOnline/problem.php?pid=28代码如下：[java] view plai

大数阶乘时间限制：3000 ms  |  内存限制：65535 KB难度：3描述我们都知道如何计算一个数的阶乘，可是，如果这个数很大呢，我们该如何去计算它并输出它？输入输入一个整数m(0输出输出m的阶乘，并在输出结束之后输入一个换行符样例输入50样例输出30414093201713378043612608166064768844

Hat's FibonacciTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7857    Accepted Submission(s): 2553Problem DescriptionA Fibonacci se

N!Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 53785    Accepted Submission(s): 15217Problem DescriptionGiven an integer N(0 ≤ N ≤ 1

Problem DescriptionThe digital root of a positive integer is found by summing the digits of the integer. If the resulting value is a single digit then that digit is the digital root. If the resultin

Big NumberTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 26337    Accepted Submission(s): 11979Problem DescriptionIn many applicati

/*大菲波数Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 11168    Accepted Submission(s): 3782Problem DescriptionFibonacci数列，定义如下：f(1

/*A + B Problem IITime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 210207    Accepted Submission(s): 40482Problem DescriptionI have a

/*循环多少次？Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2839    Accepted Submission(s): 1028Problem Description  我们知道，在编程中，我们时常需要考虑到

Problem DescriptionYou will be given a string which only contains ‘1’; You can merge two adjacent ‘1’ to be ‘2’, or leave the ‘1’ there. Surly, you may get many different results. For example, given

题目链接：http://poj.org/problem?id=2506递推：f [0] =1；//不要问为什么，递推就是这么推的f[1] = 1； f[2] =3； f[n]= f[n-2]*2 + f[n-1] ；大数：二维数组 #include#include #includeusing namespace std;int dp[255][8