light oj 1032 - Fast Bit Calculations-优快云博客

本文介绍了一种使用数位动态规划解决特定位运算问题的方法，该问题旨在计算从0到N的所有整数中相邻1出现的总次数。通过定义状态转移方程，实现了高效求解。

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1032 - Fast Bit Calculations

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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

Output for Sample Input

2147483647

Case 1: 0

Case 2: 2

Case 3: 12

Case 4: 13

Case 5: 13

Case 6: 14

Case 7: 16106127360

数位DP.

dp[pos][sum][st]:当前位数为pos时，sum为“11”有多少个，st为前一位为0或1；

#include<iostream>
using namespace std;
#include<string.h>
long long dp[50][50][2];
int dight[50];
long long dfs(int pos,int sum,int st,bool flag)
{  int i;
   long long ans=0;
    if(pos==-1)return sum;
    if(!flag&&dp[pos][sum][st]!=-1)return dp[pos][sum][st];
    int u=flag?dight[pos]:1;
    for(i=0;i<=u;i++)
        ans=ans+dfs(pos-1,sum+(st&&i==1),i,flag&&i==u);
    if(!flag)dp[pos][sum][st]=ans;
    return ans;
}
long long fun(long long n)
{
    int len=0;
    while(n)
    {
        dight[len++]=n%2;
        n=n/2;
    }
    dfs(len-1,0,0,1);
}
int main()
{
    int t,i;
    long long n;
    cin>>t;
    for(i=1;i<=t;i++)
    { memset(dp,-1,sizeof(dp));
      cin>>n;
      cout<<"Case "<<i<<":"<<" "<<fun(n)<<endl;
    }
    return 0;
}