Divided Land Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 123    Accepted Submission(s): 64 Problem Description It’s time to fight the local despots and redistribute the land. There is a rect…

题意  给你两个二进制数m,n   求他们的最大公约数  用二进制表示  0<m,n<2^1000 先把二进制转换为十进制  求出最大公约数  再把结果转换为二进制  数比較大要用到大数 import java.util.*; import java.math.*; public class wl6_9 { static BigInteger two = BigInteger.valueOf(2), one = BigInteger.ONE, zero = BigInteger.ZERO; s…

题目要求就是做求两个二进制数的gcd,如果是用java的话,这题很简单.但也可以用C++做,只能先给自己留下这个坑了,还在研究c++的做法. import java.math.BigInteger; import java.util.Scanner; /** * Created by emerald on 8/14/15. * */ public class Main { public static void main(String []args) { Scanner in = new Scan…

Divided Land Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 56    Accepted Submission(s): 27 Problem Description    It’s time to fight the local despots and redistribute the land. There is a r…

http://acm.hdu.edu.cn/showproblem.php?pid=5050 题目大意: 给定一个矩形的长和宽,把这个矩形分成若干相等的正方形,没有剩余.求正方形的边长最长是多少. 解题思路: 这道题是“pick定理”的一个变种(不知道是pick定理,也可以退出结论).由定理是求矩形的长和宽最大公约数.但是这道题,输入的数是二进制, 输出也是二进制,二进制数的大小为2^1000,所以是大数的算法.java有大数的处理. 1 import java.math.BigInteger;…

It's time to fight the local despots and redistribute the land. There is a rectangular piece of land granted from the government, whose length and width are both in binary form. As the mayor, you must segment the land into multiple squares of equal s…

这个函数是我无意中看到的很不错,很给力,我喜欢 是用于求最小公约数的 简单的描述就是,记gcd(a,b)表示非负整数a,b的最大公因数,那么:gcd(a,b)=gcd(b,a%b)或者gcd(a,0)=gcd(0,a)=a 请看代码 int gcd(int a,int b){ if(a==0) return b; if(b==0) return a; return gcd(b,a%b);} 例题 链接 http://acm.hdu.edu.cn/showproblem.php?pid=1108…

话不多说,日常一水题,水水更健康!┗|｀O′|┛ 嗷~~ a/b + c/d Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 14345    Accepted Submission(s): 7470 Problem Description 给你2个分数,求他们的和,并要求和为最简形式.   Input 输入首先包含一个正整数T(T<=…

Euclid求最大公约数算法 #include <stdio.h> int gcd(int x,int y){ while(x!=y){ if(x>y) x=x-y; else y=y-x; } return x; } int main(int argc, const char *argv[]) { if(3!=argc){ printf("Usage:<a,out> num1 num2\n"); return -1; } printf("%d\…

#欧几里得求最大公约数 #!/usr/bin/env python #coding -*- utf:8 -*- #iteration def gcd(a,b): if b==0: return a else: return gcd(b, remainder(a, b)) #此方法仅仅书用于a和b都为正数 def gcd_1(a,b): while(b>0): rem = remainder(a,b) a = b b = rem return a def remainder(x,y): retur…