lightoj 1032 - 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	Output for Sample Input
7 0 6 15 20 21 22 2147483647	Case 1: 0 Case 2: 2 Case 3: 12 Case 4: 13 Case 5: 13 Case 6: 14 Case 7: 16106127360

题意：给你一个数n，问你在二进制下1～n中共有几对11。

一道数位dp。

dp[len][count][pre] len表示当前位数，count表示1～len位有几对“11”，pre表示上一位是什么。

#include <iostream>
#include <cstring>
using namespace std;
typedef long long LL;
LL dp[100][100][2];
LL dig[100];
LL dfs(int x , int pre , int flag , int count)
{
    if(x == -1)
        return count;
    if(dp[x][count][pre] != -1 && !flag)
        return dp[x][count][pre];
    int n = flag ? dig[x] : 1;
    LL sum = 0;
    for(int i = 0 ; i <= n ; i++)
    {
        if(pre == 1 && i == 1)
            sum += dfs(x - 1 , i , flag && i == n , count + 1);
        else
        {
            sum += dfs(x - 1 , i , flag && i == n , count);
        }
    }
    if(!flag)
        dp[x][count][pre] = sum;
    return sum;
}
LL Gets(int x)
{
    memset(dig , 0 , sizeof(dig));
    int len = 0;
    while(x)
    {
        dig[len++] = (x & 1);
        x >>= 1;
    }
    return dfs(len - 1 , 0 , 1 , 0);
}
int main()
{
    int t;
    cin >> t;
    int ans = 0;
    while(t--)
    {
        ans++;
        int n;
        cin >> n;
        memset(dp , -1 , sizeof(dp));
        cout << "Case " << ans << ": " << Gets(n) << endl;
    }
    return 0;
}