Light oj 1032 - Fast Bit Calculations（数位dp）

最新推荐文章于 2021-08-20 10:10:36 发布

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本文介绍了一个算法问题，即计算从0到N的所有整数中相邻位为1的情况总数。通过递归方法和动态规划实现，文章提供了一段C++代码示例，用于解决这个问题。

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

PROBLEM SETTER: MOHIUL ALAM PRINCE

SPECIAL THANKS: JANE ALAM JAN (MODIFIED DESCRIPTION, DATASET)

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-8
typedef long long ll;

using namespace std;


#define INF 0x3f3f3f3f
#define N 33

ll dp[N][N][2];
int bit[N];

ll dfs(int pos,int cur,int pre,bool bound)
{
     if(pos==-1) return cur;
     if(!bound&&dp[pos][cur][pre]>=0) return dp[pos][cur][pre];
     ll ans=0;
     int up=bound ? bit[pos]:1;
     for(int i=0;i<=up;i++)
     {
         if(i&1)
            ans+=dfs(pos-1,cur+pre,i,bound&&i==up);
         else
            ans+=dfs(pos-1,cur,i,bound&&i==up);
     }
     if(!bound) dp[pos][cur][pre]=ans;
     return ans;
}

ll fdd(int n)
{
    int len=0;
    while(n)
    {
        bit[len++]=n%2;
        n/=2;
    }
    return dfs(len-1,0,0,true);
}

int main()
{
    int i,j,t,ca=0,n;
    memset(dp,-1,sizeof(dp));
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        printf("Case %d: %lld\n",++ca,fdd(n));
    }
    return 0;
}