Codeforces Round #458C DP
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.
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 a single integer — the number of special numbers not greater than n, modulo 109 + 7.
110
2
3
111111011
2
169
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).
代码:
1 //#include "bits/stdc++.h" 2 #include "cstdio" 3 #include "map" 4 #include "set" 5 #include "cmath" 6 #include "queue" 7 #include "vector" 8 #include "string" 9 #include "cstring" 10 #include "time.h" 11 #include "iostream" 12 #include "stdlib.h" 13 #include "algorithm" 14 #define db double 15 #define ll long long 16 //#define vec vector<ll> 17 #define Mt vector<vec> 18 #define ci(x) scanf("%d",&x) 19 #define cd(x) scanf("%lf",&x) 20 #define cl(x) scanf("%lld",&x) 21 #define pi(x) printf("%d\n",x) 22 #define pd(x) printf("%f\n",x) 23 #define pl(x) printf("%lld\n",x) 24 #define rep(i, x, y) for(int i=x;i<=y;i++) 25 const int N = 1e3 + 9; 26 const int mod = 1e9 + 7; 27 const int MOD = mod - 1; 28 const db eps = 1e-10; 29 const db PI = acos(-1.0); 30 using namespace std; 31 32 int f[N],a[N],C[N][N]; 33 int n,m,ans,tot; 34 char str[N]; 35 36 int calc(int x) 37 { 38 int res=0; 39 while(x){if(x&1)res++;x>>=1;} 40 return res; 41 } 42 43 void work(int x) 44 { 45 int cnt=0; 46 for(int i=n;i>=1;i--)//x个1放在n个位置上 47 { 48 if(a[i])//当原数字对应位为1时,可以放1与0。 49 { 50 if(x>=cnt) ans=(ans+C[i-1][x-cnt])%mod;// 此位不放1的种数 51 cnt++;//之后此位放1 52 } 53 } 54 if(x==tot) ans=(ans+1)%mod; 55 } 56 57 int main() 58 { 59 scanf("%s%d",str+1,&m);n=strlen(str+1); 60 for(int i=1;i<=n;i++) a[i]=str[i]-'0',tot+=a[i]; 61 reverse(a+1,a+1+n); 62 for(int i=0;i<=n;i++) 63 { 64 C[i][0]=1; 65 for(int j=1;j<=i;j++) C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod; 66 } 67 f[1]=0; 68 for(int i=2;i<=1000;i++) f[i]=f[calc(i)]+1; 69 if(m==0) puts("1"); 70 else 71 { 72 for(int i=1;i<=1000;i++) if(f[i]==m-1) work(i); 73 if(m==1) ans--; 74 printf("%d\n",(ans+mod)%mod); 75 } 76 return 0; 77 }