lightoj1032_数位dp

Posted 2020-08-13 _lm

题目链接：http://lightoj.com/volume_showproblem.php?problem=1032

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
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

 思路:dp[len][pre][con]表示到当前位置，前一位是pre时，前面有多少个11的答案是多少

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <cstdio>
 6 #include <vector>
 7 #include <ctime>
 8 #include <queue>
 9 #include <list>
10 #include <set>
11 #include <map>
12 using namespace std;
13 #define INF 0x3f3f3f3f
14 typedef long long LL;
15 
16 int bit[50];
17 LL dp[50][2][50];
18 LL dfs(int len, int pre, int con, int flag)
19 {
20     if(len == 0)
21         return con;
22     if(dp[len][pre][con] >= 0 && flag)
23         return dp[len][pre][con];
24     int te = flag ? 1 : bit[len];
25     LL sum = 0;
26     for(int i = 0; i <= te; i++)
27     {
28         if(pre == 1 && i == 1)
29             sum += dfs(len-1, i, con+1, flag || i < te);
30         else
31             sum += dfs(len-1, i, con, flag || i < te);
32     }
33     if(flag)
34         dp[len][pre][con] = sum;
35     return sum;
36 }
37 int main()
38 {
39     int t, n;
40     scanf("%d", &t);
41     memset(dp, -1, sizeof(dp));
42     for(int ca = 1; ca <= t; ca++)
43     {
44         scanf("%d", &n);
45         int len = 0;
46         while(n)
47         {
48             bit[++len] = n % 2;
49             n >>= 1;
50         }
51         printf("Case %d: %lld\\n", ca, dfs(len, 0, 0, 0));
52     }
53     return 0;
54 }

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