Discription
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 1310?=?11012, 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 109?+?7.
Input
The first line contains integer n (1?≤?n?<?21000).
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 109?+?7.
Example
Input
1102
Output
3
Input
1111110112
Output
169
Note
In the first sample, the three special numbers are 3, 5 and 6. They get reduced to 2 in one operation (since there are two set bits in each of 3, 5 and 6) and then to 1 in one more operation (since there is only one set bit in 2).
最简单的数位dp类型，多加一维0\1表示是否贴上界(当初自己想的方法没想到还挺靠谱，，，基本上用了的题都对了)。
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<vector>
#define ll long long
#define maxn 1010
#define ha 1000000007
using namespace std;
char s[maxn];
int n,m,c[maxn];
int num[maxn],k,ans;
int f[maxn][maxn][2];
//f[i][j][0/1]表示前i位中和为j，且是否贴上界的方案数
inline int add(int x,int y){
x+=y;
if(x>=ha) x-=ha;
return x;
}
inline void init(){
num[0]=0;
for(int i=1;i<=1005;i++) num[i]=num[i^(i&-i)]+1;
c[1]=0,c[0]=123456;
for(int i=2;i<=1005;i++){
c[i]=c[num[i]]+1;
//c[i]<5,虽然可能并没有什么卵用
}
}
inline void solve(){
n=strlen(s+1);
//一开始是贴上界的，因为之前的位都是0。
f[0][0][1]=1;
for(int i=1;i<=n;i++) if(s[i]==‘1‘){
//这一位是1的话是不可能选的数1个数为0且贴上界
f[i][0][0]=add(f[i-1][0][0],f[i-1][0][1]);
f[i][0][1]=0;
for(int j=1;j<=i;j++){
//不贴上界可能是 这一位为0且之前是否贴上界任意 或者 这一位为1且之前不贴上界
f[i][j][0]=add(add(f[i-1][j][0],f[i-1][j][1]),f[i-1][j-1][0]);
//贴上界只能是之前贴上界且当前位是1
f[i][j][1]=f[i-1][j-1][1];
}
}
else{
f[i][0][0]=f[i-1][0][0];
f[i][0][1]=f[i-1][0][1];
for(int j=1;j<=i;j++){
//不贴上界的话只能 这一位为0且之前不贴上界 或者 这一位为1且之前不贴上界
f[i][j][0]=add(f[i-1][j][0],f[i-1][j-1][0]);
//之所以没有f[i-1][j-1][1]是因为这一位是1的话是不可能贴上界的
f[i][j][1]=f[i-1][j][1];
}
}
ans=0;
for(int i=0;i<=n;i++) if(c[i]==k-1) ans=add(ans,add(f[n][i][0],f[n][i][1]));
//这样枚举<=n的数的1的个数会有一个漏洞，那就是1将会被算到k==1的里面去，所以要特判一下
if(k==0) ans++;
else if(k==1) ans--;
}
int main(){
init();
scanf("%s",s+1);
scanf("%d",&k);
solve();
printf("%d\n",ans);
return 0;
}
原文地址：https://www.cnblogs.com/JYYHH/p/8419456.html