HDU 4588 Count The Carries

Count TheCarries

Time Limit: 4000/2000 MS(Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 207    Accepted Submission(s): 97

Problem Description

One day, Implusgets interested in binary addition and binary carry. He will transfer alldecimal 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 inorder. For example, from 1 to 3 (decimal digit), he will firstly calculate 01(1)+10 (2), get 11，then calculate11+11 (3)，lastly 110 (binarydigit), we can find that in the total process, only 2 binary carries happen. Hewants to find out that quickly. Given a and b in decimal, we transfer intobinary digits and use Implus's addition algorithm, how many carries are there？

 

 

Input

Two integers a, b（0<=a<=b<1000000000), about 100000cases, end with EOF.

 

 

Output

One answer perline.

 

 

Sample Input

1 2

1 3

1 4

1 6

 

 

Sample Output

0

2

3

6

 

看了半天也不知道怎么解。写了很久发现自己发明的算法超时，根本算不出来。

看了 别人写的题解，这才懂了。

比如1 + 2 + 3 + 4：

    1

  10

  11

100

——

122

把各位的1的数目统计出来，和十进制一样计算，二进制可以存进int  num[70]数组中，从num[0]往num[69]算，也就是从低位往高位算。

然后对122进行进位，122>130>210>1010  算的过程中整除2得到进位的次数，并且加到前一位，如此循环。

 

 

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std;

__int64 num1[70],num2[70];

void calc(__int64 num[],__int64 n){
    int tmp=1;
    for(int i=0;i<63;i++)
        num[i]=0;
    for(int i=0;i<63;i++){
        if(n<=0) break;
        tmp*=2;
        num[i]=(n-n%tmp)/2;
        if(n%tmp>=tmp/2)
            num[i]+=tmp/2;
        else
            num[i]+=n%tmp;
        n=n-tmp/2;
    }
}

int main(){
    __int64 a,b;
    while(~scanf("%I64d %I64d",&a,&b)){
        __int64 ans=0;
        calc(num1,a-1);
        calc(num2,b);
        for(int i=0;i<63;i++)
            num2[i]=num2[i]-num1[i];
        for(int i=0;i<62;i++){
            ans+=num2[i]/2;
            num2[i+1]+=num2[i]/2;
        }
        printf("%I64d\n",ans);
    }
    return 0;
}

代码引用自(u011523796)    已经AC 没有问题。