2017 CCPC 秦皇岛 & ZOJ 3987 - Numbers （贪心+大数）

最新推荐文章于 2022-11-25 16:09:51 发布

原创最新推荐文章于 2022-11-25 16:09:51 发布 · 505 阅读

0 ·

CC 4.0 BY-SA版权

贪心同时被 3 个专栏收录

32 篇文章

订阅专栏

ZOJ

21 篇文章

订阅专栏

大数

7 篇文章

订阅专栏

本文介绍了一种使用Java处理大数的贪心算法方法，该方法应用于解决将一个非负整数拆分为多个非负整数的问题，并最小化这些整数的按位或运算结果。文章通过一个具体的竞赛题目来展示算法实现细节。

Numbers

Time Limit: 2 Seconds Memory Limit: 65536 KB

DreamGrid has a nonnegative integer . He would like to divide into nonnegative integers and minimizes their bitwise or (i.e. and should be as small as possible).

Input

There are multiple test cases. The first line of input contains an integer , indicating the number of test cases. For each test case:

The first line contains two integers and ().

It is guaranteed that the sum of the length of does not exceed .

Output

For each test case, output an integer denoting the minimum value of their bitwise or.

Sample Input

Sample Output

Author: LIN, Xi
Source: The 2017 China Collegiate Programming Contest, Qinhuangdao Site

POINT：

第一次用java写大数，好方便啊！

这题简单的贪心就行了。

从高位开始判断，判断这一位的0可以不可以给：

可以给的状态是假设这一位后面的数全是1111，这1111*m不小于当前要填满的n，那么这个0就可以放。

import java.math.BigInteger;
import java.util.Scanner;
import java.math.*;


public class Main {
	public static void main(String[] args) {
		Scanner cin = new Scanner(System.in);  
		int T;
		while(cin.hasNext()){  
			T=cin.nextInt();
			while(T!=0) {
				BigInteger n = cin.nextBigInteger();
				BigInteger m = cin.nextBigInteger();
				BigInteger now = BigInteger.valueOf (1);
				BigInteger er =BigInteger.valueOf(2);
				BigInteger sum = m.multiply(now);
				int k=1;
				while(sum.compareTo(n)==-1) {
					k++;
					now=now.multiply(er);
					sum=sum.add(now.multiply(m));
				}
				BigInteger ans = BigInteger.ZERO;
			//	System.out.println(now);
				while(n.compareTo(BigInteger.ZERO)!=0) {
					BigInteger num = n.divide(now);
					if(num.compareTo(m)==1) {
						num=m;
					}
					BigInteger now1 = now.subtract(BigInteger.ONE);
					if(now1.multiply(m).compareTo(n)!=-1) {
						now=now.divide(er);
						continue;
						
					}
					ans=ans.add(now);
					n=n.subtract(num.multiply(now));
					now=now.divide(er);
				}
				System.out.println(ans);
				T--;
			}
        }
	}

}