Lightoj1032——Fast Bit Calculations（数位dp）

A bit is a binary digit, taking a logical value of either 1 or 0 (also referred to as “true” or “false” respectively). And every decimal number has a binary representation which is actually a series of bits. If a bit of a number is 1 and its next bit is also 1 then we can say that the number has a 1 adjacent bit. And you have to find out how many times this scenario occurs for all numbers up to N.

Examples:

Number         Binary          Adjacent Bits

12                    1100                        1
15                    1111                        3
27                    11011                      2

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer N (0 ≤ N < 231).

Output

For each test case, print the case number and the summation of all adjacent bits from 0 to N.

Sample Input

7

0

6

15

20

21

22

2147483647

Output for Sample Input

Case 1: 0

Case 2: 2

Case 3: 12

Case 4: 13

Case 5: 13

Case 6: 14

Case 7: 16106127360

转换成二进制dp

#include <iostream>
#include <cstring>
#include <string>
#include <vector>
#include <queue>
#include <cstdio>
#include <set>
#include <cmath>
#include <map>
#include <algorithm>
#define INF 0x3f3f3f3f
#define MAXN 10000005
#define Mod 10001
using namespace std;
int dight[40];
long long dp[40][2][40];
long long dfs(int pos,int s,bool limit,int sum)
{
if(pos==0)
return sum;
if(!limit&&dp[pos][s][sum]!=-1)
return dp[pos][s][sum];
int end;
long long ret=0;
if(limit)
end=dight[pos];
else
end=1;
for(int d=0; d<=end; ++d)
{
if(s)
{
if(d==1)
ret+=dfs(pos-1,1,limit&&d==end,sum+1);
else
ret+=dfs(pos-1,0,limit&&d==end,sum);
}
else
{
if(d==1)
ret+=dfs(pos-1,1,limit&&d==end,sum);
else
ret+=dfs(pos-1,0,limit&&d==end,sum);
}
}
if(!limit)
dp[pos][s][sum]=ret;
return ret;
}
long long solve(long long a)
{
memset(dight,0,sizeof(dight));
int cnt=1;
while(a!=0)
{
dight[cnt++]=a%2;
a/=2;
}
return dfs(cnt-1,0,1,0);
}
int main()
{
memset(dp,-1,sizeof(dp));
int t,cnt=1;
scanf("%d",&t);
while(t--)
{
long long x;
scanf("%lld",&x);
printf("Case %d: %lld\n",cnt++,solve(x));
}
return 0;
}