UVALive 4864 数位dp

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 }

 

转载于:https://www.cnblogs.com/nuoyan2010/archive/2013/05/08/3067186.html