Codeforces Gym 100231L Intervals 数位DP

Intervals

题目连接：

http://codeforces.com/gym/100231/attachments

Description

Start with an integer, N0, which is greater than 0. Let N1 be the number of ones in the binary representation of N0. So, if N0 = 27, N1 = 4. For all i > 0, let Ni be the number of ones in the binary
representation of Ni−1. This sequence will always converge to one. For any starting number, N0, let K be the minimum value of i ≥ 0 for which Ni = 1. For example, if N0 = 31, then N1 = 5, N2 = 2, N3 = 1, so K = 3. Given a range of consecutive numbers, and a value X, how many numbers in the range have a K value equal to X?

Input

There will be several test cases in the input. Each test case will consist of three integers on a single line:
l, r, X, where l and r (1 ≤ l ≤ r ≤ 1018) are the lower and upper limits of a range of integers, and X
(0 ≤ X ≤ 10) is the target value for K. The input will end with a line with three 0s.

Output

For each test case, output a single integer, representing the number of integers in the range from l to
r (inclusive) which have a K value equal to X in the input. Print each integer on its own line with no
spaces. Do not print any blank lines between answers.

Sample Input

31 31 3

31 31 1

27 31 1

27 31 2

1023 1025 1

1023 1025 2

0 0 0

Sample Output

1

0

0

3

1

1

Hint

题意

首先给你Ni的定义，表示第几轮的时候，这个数是多少，Ni = Ni-1二进制表示下的1的个数

k 表示第几步的时候，Ni = 1

给你l,r,x

问你在l,r区间内，k等于x的数有多少个

题解：

我们首先预处理vis[i]表示有i个1的时候的步数，这个用dp很容易解决

然后我们就可以数位dp去做了，做[1,x]里面二进制数为k个的数量

注意特判1的情况，比较麻烦

代码

#include<bits/stdc++.h>
using namespace std;

long long l,r,t;
int vis[100];
long long ans = 0;
int getone(long long x)
{
    int c=0;
    while(x>0)
    {
        if((x&1)==1)
            c++;
        x>>=1;
    }
    return c;
}
long long f[70][70];
void init()
{
   memset(f,0,sizeof(f));
   f[0][0] = 1LL;
   for(int i=1;i<=62;i++)
   {
      f[i][0] = 1LL;
      for(int j=1;j<=i;j++)
      {
         f[i][j] = f[i-1][j-1] + f[i-1][j];
      }
   }
}
long long calc(long long x,int k)
{
   int tot = 0;
   long long ans = 0;
   for(long long i=62;i>0;i--)
   {
      if(x&(1LL<<i))
      {
         tot++;
         if(tot>k) break;
         x ^= (1LL<<i);
      }
      if((1LL<<(i-1LL))<=x)
      {
          if(k>=tot)
            ans += f[i-1][k-tot];
      }
   }
   if(tot + x == k) ans++;
   return ans;
}
long long solve(long long limit,int x)
{
    ans=0;
    for(int i=1;i<=61;i++)
        if(vis[i]==x)
        {
            if(i==1)
                ans--;
            ans+=calc(limit,i);
        }
    return ans;
}
int main()
{
    init();
    vis[1]=1;
    for(int i=2;i<=61;i++)
        vis[i]=vis[getone(i)]+1;
    while(scanf("%lld%lld%d",&l,&r,&t)!=EOF)
    {
        if(l==0&&r==0&&t==0)return 0;
        if(t==0)
        {
            if(l==1)
                printf("1
");
            else
                printf("0
");
            continue;
        }
        if(t==1)
        {
            if(l==1)
                printf("%lld
",solve(r,t)-solve(l-1,t)-1);
            else
                printf("%lld
",solve(r,t)-solve(l-1,t));
        }
        else
            printf("%lld
",solve(r,t)-solve(l-1,t));
    }
}