版权声明：抱最大的希望，为最大的努力，做最坏的打算。 https://blog.csdn.net/qq_37748451/article/details/86588695
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
Sample Output
Case 1: 0
Case 2: 2
Case 3: 12
Case 4: 13
Case 5: 13
Case 6: 14
Case 7: 16106127360题意：输入n，要求从0到n的所有数的二进制数里面总共有多少“11”。 求每个数出现11的次数之和
思路： 一看就是数位dp的标准题目，问题就在于如何去写状态转移方程
将输入的数转化为二进制之后进行数位dp。
dp[i][j][k]表示i位数前一位是j（由于二进制只可能是0或1）已经存在k个11的总数。
直接上代码去理解吧
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-6
#define MOD 1000000007
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
ll dp[40][2][40];//i位数，前一位是j，已经存在k个11的总数
int digit[40];
ll dfs(int pos,int pre,int st,bool limit)
{
if(pos==0)return st;
if(!limit&&dp[pos][pre][st]!=-1)return dp[pos][pre][st];
ll ans=0;
int end=limit?digit[pos]:1;
for(int i=0; i<=end; i++)
ans+=dfs(pos-1,i,st+(pre&&i),limit&&(i==end));
if(!limit)
dp[pos][pre][st]=ans;
return ans;
}
ll get(int x)
{
int bj=0;
while(x)
digit[++bj]=x%2,x/=2;
return dfs(bj,0,0,1);
}
int main()
{
memset(dp,-1,sizeof(dp));
int t,o=1;
scanf("%d",&t);
while(t--)
{
int n;
scanf("%d",&n);
printf("Case %d: %lld\n",o++,get(n));
}
return 0;
}