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?
64位同时为1时即转换为了64以下的数转换到1的步数，所以首先打个表，之后枚举数位之和就ok了。
View Code
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<cstdlib> 5 using std::cout; 6 using std::cin; 7 using std::endl; 8 typedef long long LL; 9 int const N = 64; 10 int step[N],bit[N+10],ln; 11 LL lo,hi; 12 int x; 13 LL dp[N][N][2]; 14 int getsum1(int n) 15 { 16 int sum=0; 17 int num=n; 18 for(;n;sum+=n%2,n>>=1); 19 if(sum==1)return 1; 20 if(step[sum])return step[sum]+1; 21 step[num]=getsum1(sum)+1; 22 } 23 void pre() 24 { 25 step[1]=0; 26 for(int i=2;i<=64;i++) 27 { 28 if(!step[i]) 29 step[i]=getsum1(i); 30 } 31 } 32 LL getsum2(int t,int pre,int rest,int limit) 33 { 34 if(rest<0)return 0; 35 if(!t)return (rest==0); 36 int up=(limit?bit[t]:1); 37 LL ans=0; 38 if(!limit&&dp[t][rest][pre]!=-1)return dp[t][rest][pre]; 39 for(int i=0;i<=up;i++) 40 { 41 ans+=getsum2(t-1,i,rest-i,limit&&i==up); 42 } 43 if(!limit&&dp[t][rest][pre]==-1)dp[t][rest][pre]=ans; 44 return ans; 45 } 46 LL getsum3(LL n) 47 { 48 if(n<=0)return 0; 49 for(ln=0;n;bit[++ln]=n%2,n>>=1); 50 LL ans=0; 51 if(x==0) 52 return 1; 53 if(x==step[1]+1) 54 ans+=getsum2(ln,0,1,1)-1; 55 for(int i=2;i<=ln;i++) 56 if(x==step[i]+1) 57 ans+=getsum2(ln,0,i,1); 58 return ans; 59 } 60 int main() 61 { 62 pre(); 63 memset(dp,-1,sizeof(dp)); 64 while(scanf("%lld %lld",&lo,&hi)) 65 { 66 scanf("%d",&x); 67 if(!(x+lo+hi))break; 68 printf("%lld\n",getsum3(hi)-getsum3(lo-1)); 69 } 70 return 0; 71 }