题目地址:http://acm.hdu.edu.cn/showproblem.php?pid=4588
思路:
0 0000
1 0001
2 0010
3 0011
4 0100
5 0101
6 0110
7 0111
8 1000
9 1001
10 1010
观察可知,第i位有(1<<i)个0,与(1<<i)个1,按照长度为(1<<(i+1))长度循环。则对于数n,可以求出1到n中各位1的个数num2[i],则进位次数为sum(num[i] / 2),注意每次向高位进位(num2[i+1]+=num2[i] / 2)。
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long LL;
LL num1[65],num2[65];
void solve(LL num[],LL n)
{
for(int i=0;i<63;i++) num[i]=0;
n++;
for(int i=0;i<63;i++)
{
LL tmp1=(1LL)<<(i+1);
LL tmp2=(1LL)<<i;
num[i]+=n/tmp1*(1<<i);
if(n%tmp1>tmp2)num[i]+=n%tmp1-tmp2;
}
/*for(int i=0;i<63;i++)
cout<<i<<" "<<num[i]<<endl;*/
}
int main()
{
LL a,b;
//solve(num1,20);
//solve(num2,6);
while(scanf("%I64d%I64d",&a,&b)==2)
{
solve(num1,a-1);
solve(num2,b);
for(int i=0;i<63;i++) num2[i]-=num1[i];
LL ans=0;
for(int i=0;i<63;i++)
{
ans+=num2[i]/2;
num2[i+1]+=num2[i]/2;
}
printf("%I64d\n",ans);
}
return 0;
}