2017CCPC秦皇岛G ZOJ 3987Numbers（大数+贪心）

Time Limit: 2 Seconds      Memory Limit: 65536 KB

DreamGrid has a nonnegative integer n . He would like to divide n into
m nonnegative integers a1,a2,...am and minimizes their bitwise
or (i.e.a1+a2+...+am=n  and a1 OR a2 OR a3...OR am 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 n and  m( 0<=n<=1e1000 1<=m<=1e100).

It is guaranteed that the sum of the length of does not exceed 20000

Output

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

Sample Input

5
3 1
3 2
3 3
10000 5
1244 10

Sample Output

3
3
1
2000
125

Author: LIN, Xi

Source: The 2017 China Collegiate Programming Contest, Qinhuangdao Site

/*
* 题意：给你一个1e1000的数，让你分成最多1e100位，然后互相进行或运算，得到的值最小
*
* 思路：让高位最小然后向低位贪心,判断当前为能否为0，能为0的条件是,后面几位都是1,并且有m个,加起来如果大于当前值的话
*   那么这位就可以为0,否则只能为1,既然为1了,那么尽量多填1,这样保证了结果最优
* */
import java.math.BigInteger;
import java.util.Scanner;

public class Main {

public static void main(String[] args) {
// write your code here
Scanner in=new Scanner(System.in);
int t;
BigInteger n,m,x,sum;
BigInteger [] pre=new BigInteger[4005];//表示后缀和
String s;
x=new BigInteger("1");
pre[0]=new BigInteger("0");
for(int i=1;i<4005;i++){
pre[i]=pre[i-1].add(x);
x=x.multiply(new BigInteger("2"));
}
t=in.nextInt();
for(int ca=0;ca<t;ca++){
n=in.nextBigInteger();
m=in.nextBigInteger();
sum=n;
int tol=0;
while(n.compareTo(new BigInteger("0"))==1){
tol++;
n=n.divide(new BigInteger("2"));
}
x=new BigInteger("0");
//判断这一位的累加和能不能用后边的所有空间放下，能的话就转移到下一位，不能的话就把这一位都填满
for(int i=tol;i>=1;i--){//枚举每位
x=x.multiply(new BigInteger("2"));
int ok=sum.compareTo(pre[i - 1].multiply(m));
if ( ok<=0) {//能用剩余的空间放下
continue;
} else {//放不下
x = x.add(new BigInteger("1"));
BigInteger cur = pre[i].subtract(pre[i - 1]);
BigInteger cnt = sum.divide(cur);
if (cnt.compareTo(m) >= 0) {
sum = sum.subtract(cur.multiply(m));
} else {
sum = sum.subtract(cur.multiply(cnt));
}
}
}
System.out.println(x);
}
}
}

//package aax;

import java.math.BigInteger;

import java.util.Scanner;

public class Main{
public static void main(String[] args){
Scanner in=new Scanner(System.in);
int t;
BigInteger n, m, x, sum;
BigInteger []pre=new BigInteger[4005];
String s;
x=new BigInteger("1");
pre[0]=new BigInteger("0");
for(int i=1;i<4005;i++)
{
pre[i]=pre[i-1].add(x);
x=x.multiply(new BigInteger("2"));
}
t=in.nextInt();
for(int cas=0;cas<t;cas++)
{
n=in.nextBigInteger();
m=in.nextBigInteger();
sum=n;
int tol=0;
while(n.compareTo(new BigInteger("0"))==1)
{
tol++;
n=n.divide(new BigInteger("2"));
}
x=new BigInteger("0");
for(int i=tol;i>=1;i--)
{
x=x.multiply(new BigInteger("2"));
int ok=sum.compareTo(pre[i-1].multiply(m));
if(ok<=0)
{
continue;
}
else
{
x=x.add(new BigInteger("1"));
BigInteger cur=pre[i].subtract(pre[i-1]);
BigInteger cnt=sum.divide(cur);
if(cnt.compareTo(m)>=0)
{
sum=sum.subtract(cur.multiply(m));
}
else
{
sum=sum.subtract(cur.multiply(cnt));
}
}
}
System.out.println(x);
}
}
}