hnu Counting ones 统计1-n 二进制中1的个数

 

Counting ones

Time Limit: 1000ms, Special Time Limit:2500ms, Memory Limit:65536KB

Total submit users: 18, Accepted users: 16

Problem 13030 : No special judgement

Problem description

Carl is right now the happiest child in the world: he has just learned this morning what the bi- nary system is. He learned, for instance, that the binary representation of a positive integer k is a string a_na_n-1 ... a₁a₀ where each ai is a binary digit 0 or 1, starting with a_n = 1, and such that k = Σa_i × 2ⁱ(i from 0 to n). It is really nice to see him turning decimal numbers into binary numbers, and then adding and even multiplying them.
Caesar is Carl's older brother, and he just can't stand to see his little brother so happy. So he has prepared a challenge: "Look Carl, I have an easy question for you: I will give you two integers A and B, and you have to tell me how many 1's there are in the binary representation of all the integers from A to B, inclusive. Get ready". Carl agreed to the challenge. After a few minutes, he came back with a list of the binary representation of all the integers from 1 to 100. "Caesar, I'm ready". Caesar smiled and said: "Well, let me see, I choose A = 10¹⁵ and B = 10¹⁶. Your list will not be useful".
Carl hates loosing to his brother so he needs a better solution fast. Can you help him?

Input

A single line that contains two integers A and B (1 ≤ A ≤ B ≤ 10¹⁶).

Output

Output a line with an integer representing the total number of digits 1 in the binary representation of all the integers from A to B, inclusive.

Sample Input

Sample input 1
1000000000000000 10000000000000000

Sample input 2
2 12

Sample input 3
9007199254740992 9007199254740992

Sample Output

Sample output 1
239502115812196372

Sample output 2
21

Sample output 3
1

Problem Source

ICPC Latin American Regional 2013

 

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<cstdlib>
using namespace std;
typedef long long LL;

LL ans[65];
LL fac[65];
LL a[66],alen;
void init()
{
    fac[1]=ans[1]=1;
    fac[2]=ans[2]=2;
    for(int i=3;i<=60;i++)
    {
        fac[i] = 2*fac[i-1]+ans[i-1]-1;
        ans[i]=ans[i-1]*2;
    }
}
void get(LL n)
{
    alen = 0;
    while(n)
    {
        a[++alen]=(n&1);
        n=n>>1;
    }
}
LL solve()
{
    LL sum =0;
    LL k = 0;
    for(int i=alen;i>=1;i--)
    {
        if(a[i]==1)
        {
            sum = sum+fac[i]+k*ans[i];
            k++;
        }
    }
    return sum;
}
int main()
{
    LL n,m;
    init();
    while(scanf("%I64d%I64d",&n,&m)>0)
    {
        get(m);
        LL sum = solve();
        get(n-1);
        sum = sum-solve();
        printf("%I64d\n",sum);
    }
    return 0;
}