[Room test] numbers puzzle

小 X 同学有很强的计算能力，现在他正在玩一个游戏。
现在有一个正整数 x，每次操作他会将当前的数变为这个数写成二进制后 1 的个数
小 X 不断的进行操作，直到这个数变成 1 为止
由于小 X 的计算能力很强，他现在给出一 n 
他想知道有多少不超过 n 的正整数会在 k 次操作后变成 1 
由于答案可能很大，请对 1000000007 取模

Because the data range \ (2 ^ {1000} \) , the operation so that a maximum of only 1 of 1000

Thus the direct violence process all numbers from 1 to 1000 the number of operations becomes 1, the number of times \ (k-1 \) to be added to the digital processing stack operands denoted \ (I \)

The next task is to find the binary number which contains from 1 ~ n in \ (I \) a \ (1 \)

The method is a very simple mathematical combination, for a number, if a bit = 1,

So the answer is to contribute to the

\[\Huge C_{pos-1}^{i-cnt}\]

Wherein \ (POS \) represents the first of several is \ (1 \) , \ (CNT \) represents the number 1 in front of the

Then we can happily calculate the answer

It pits is \ (k = 1,0 \) time to pay special sentence

#include<bits/stdc++.h>
#define mod 1000000007
#define ll long long
#define lowbit(x) (x&-x)
using namespace std;

int k;
int num[1005],change;//num是把这个数字变为1的操作次数 
vector<int> v;
char c[1005];
ll ans=0;

int one_num(int x)
{
    int res=0;
    while(x) {res++;x-=lowbit(x);}
    return res;
}

ll fac[1005],invfac[1005];
ll qpow(ll n,ll k)
{
    ll res=1;
    while(k)
    {
        if(k&1) res=(res*n)%mod;
        n=(n*n)%mod;
        k>>=1;
    }
    return res;
}
void init()
{
    fac[0]=1;
    for(register int i=1;i<=1002;++i) fac[i]=(1LL*fac[i-1]*i)%mod;
}
ll inv(ll x){return qpow(x,mod-2);}
ll C(int n,int m){if(m>n||m<0) return 0;return ((1LL*fac[n]*inv(fac[m])%mod)*inv(fac[n-m]))%mod;}


int pos[1005],top=0;

int main()
{
    freopen("number.in","r",stdin);
    freopen("number.out","w",stdout);
    init();
    scanf("%s",c+1);
    scanf("%d",&k);
    if(k==0) {puts("1");return 0;}
    int len=strlen(c+1);
    num[1]=0;
    for(register int i=2;i<=len;++i)
    {
        int j=i;
        while(true)
        {
            change=one_num(j);
            num[i]++;
            if(change==1) break;
            j=change;
        }
    }
//  for(register int i=1;i<=len;++i) cout<<i<<" need: "<<num[i]<<endl;
    for(register int i=1;i<=len;++i) if(num[i]==k-1) v.push_back(i);
    
    //下面是计算 1~n 中有多少个数二进制中含 i 个 1 
//  cout<<"size:"<<v.size()<<endl;
    for(register int i=len;i>=1;--i)
        if(c[i]=='1') pos[++top]=len-i+1;
    while(!v.empty())//循环每个i 
    {
        ll i=v.back();
        ll topp=top;
        while(topp)
        {
            ans=(ans+C(pos[topp]-1,i-(top-topp)))%mod;
//          cout<<"C "<<pos[topp]-1<<" "<<i-(top-topp)<<" "<<C(pos[topp]-1,i-(top-topp))<<endl;
            topp--;
        }
        int cnt=0;
        for(register int i=1;i<=len;++i) if(c[i]=='1') cnt++;
        if(cnt==i) ans=(ans+1)%mod;
        v.pop_back();
    }
    printf("%lld\n",ans-(k==1));
    return 0;
}