HDU 4588 统计二进制加法进位次数

最新推荐文章于 2022-05-12 23:43:11 发布

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最新推荐文章于 2022-05-12 23:43:11 发布

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本文介绍了一个有趣的问题：如何统计从一个数加到另一个数的过程中，二进制表示下总的进位次数。通过分析二进制加法的特点，文章提供了一种高效的算法实现，用于解决这个问题。

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Count The Carries

Problem Description
One day, Implus gets interested in binary addition and binary carry. He will transfer all decimal digits to binary digits to make the addition. Not as clever as Gauss, to make the addition from a to b, he will add them one by one from a to b in order. For example, from 1 to 3 (decimal digit), he will firstly calculate 01 (1)+10 (2), get 11，then calculate 11+11 (3)，lastly 110 (binary digit), we can find that in the total process, only 2 binary carries happen. He wants to find out that quickly. Given a and b in decimal, we transfer into binary digits and use Implus's addition algorithm, how many carries are there？

统计用二进制从a加到b的进位次数

规律题，先统计出从a到b每一位上1的个数a[i]，打表可发现每一位上0和1出现的规律

对第i位，进位次数为a[i]/2，每次进位会使高一位上多一个1，即a[i+1]+=a[i]/2

#include <cstdio>
#include <cstring>

int aa[100],bb[100];

void gettable(int n,int m)
{
	memset(aa,0,sizeof(aa));
	memset(bb,0,sizeof(bb));
	int t=1;
	n+=1;
	m+=1;
	for(int i=0;i<64;i++)
    {
        if(t>n)
			break;
        t*=2;
        aa[i]+=(n-n%t)/2;

        if(n%t>t/2)
			aa[i]+=n%t-t/2;

    }
    t=1;
    for(int i=0;i<64;i++)
    {
        if(t>m)
			break;
        t*=2;
        bb[i]+=(m-m%t)/2;
        if(m%t>t/2)
			bb[i]+=m%t-t/2;
    }
}

int main()
{
	int a,b;
	long long ans;
	while(scanf("%d%d",&a,&b)!=EOF)
	{
		gettable(a-1,b);
		for(int i=0;i<64;i++)
		{
			bb[i]-=aa[i];
		}
		ans=0;
		for(int i=0;i<64;i++)
		{
			ans+=bb[i]/2;
			bb[i+1]+=bb[i]/2;
		}
		printf("%I64d\n",ans);
	}

}