Codeforces 914 C. Travelling Salesman and Special Numbers （dp）

Description

The Travelling Salesman spends a lot of time travelling so he tends to get bored. To pass time, he likes to perform operations on numbers. One such operation is to take a positive integer x and reduce it to the number of bits set to 1 in the binary representation of x. For example for number 13 it’s true that $13_{10} = 1101_2$ , so it has 3 bits set and 13 will be reduced to 3 in one operation.

He calls a number special if the minimum number of operations to reduce it to 1 is k.

He wants to find out how many special numbers exist which are not greater than n. Please help the Travelling Salesman, as he is about to reach his destination!

Since the answer can be large, output it modulo 10^9 + 7.

Input

The first line contains integer n (1 ≤ n < 2^1000).

The second line contains integer k (0 ≤ k ≤ 1000).

Note that n is given in its binary representation without any leading zeros.

Output

Output a single integer — the number of special numbers not greater than n, modulo 10^9 + 7.

Examples input

110
2

Examples output

3

题意

我们定义一种操作为将一个正整数变化为它二进制 1 的个数，问在小于等于给定二进制数的正整数中，有多少个数可以经过 k 次操作变化为 1 。

思路

从题中我们可以看到，一次操作可以将一个数变化为其二进制 1 的个数，于是我们设 dp[i] 表示二进制中有 i 个 1 的数需要几次变化才能到达 1 。

显然 dp[i] = dp[getNum(i)] + 1 ，其中 getNum 可以得出 i 二进制 1 的个数。

对于 dp[i] == k-1 的 i ，接下来的工作便是找出小于等于给定数且二进制 1 的个数为 i 的数的数量，直接用组合数很方便可以求得。

AC 代码

#include<bits/stdc++.h>
#define IO ios::sync_with_stdio(false);\
    cin.tie(0);\
    cout.tie(0);
using namespace std;
typedef long long LL;
const int mod = 1e9+7;
const int maxn = 1e5+10;

char str[maxn];
LL mul[maxn];
LL inv[maxn];
int dp[maxn];
int len,k;

void init()
{
    mul[0]=1;
    for(int i=1; i<maxn; i++)
        mul[i]=(mul[i-1]*i)%mod;

    inv[0]=inv[1]=1;
    for(int i=2; i<maxn; i++)
        inv[i]=(LL)(mod-mod/i)*inv[mod%i]%mod;
    for(int i=1; i<maxn; i++)
        inv[i]=(inv[i-1]*inv[i])%mod;
}

LL C(int n,int m)
{
    return mul[n]*inv[m]%mod*inv[n-m]%mod;
}

int getNum(int x)
{
    int res = 0;
    while(x)
    {
        res+=x&1;
        x>>=1;
    }
    return res;
}

int call(int x)
{
    int res = 0;
    for(int i=0; i<len; i++)
        if(str[i]-'0')
            res = (res + C(len-i-1,x--))%mod;
    return (res + (x==0))%mod;
}

void solve()
{
    len = strlen(str);
    if(k==0)
        cout<<1<<endl;
    else if(k==1)
        cout<<len-1<<endl;
    else
    {
        int ans = 0;
        for(int i=2; i<=1000; i++)
        {
            dp[i] = dp[getNum(i)] + 1;
            if(dp[i]==k-1)
                ans = (ans + call(i)) % mod;
        }
        cout<<ans<<endl;
    }
}

int main()
{
    IO;
    init();
    cin>>str>>k;
    solve();
    return 0;
}