lightoj 1032 - Fast Bit Calculations

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

如果一个二进制数，它的某一位和它后面一位都是1，那么这一位就称为adjacent bits。

问从0到N所有数的adjacent bits一共有多少个。

这个一眼就能看出来是数位DP，但是第一开始状态少想了一个，所以答案一直不对，最后加上到某位时一共有多少个adjacent bits这个状态，就得出了正确答案。

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;long long dp[40][40][2];int a[40];long long dfs(int len,int sum,int s,int fp)//到当前位一共有sum个 adjacent bits{    if(!len)        return (long long)sum;    if(!fp && dp[len][sum][s] != -1)        return dp[len][sum][s];    long long res = 0;    int n = fp ? a[len]:1;    for(int i=0;i<=n;i++)    {        if(s == 1 && i == 1)            res += dfs(len-1,sum+1,1,fp);        else            res += dfs(len-1,sum,i == 1,fp&&i==n);    }    if(!fp)        dp[len][sum][s] = res;    return res;}long long sum(int x){    int len = 0;    while(x)    {        a[++len] = x % 2;        x /= 2;    }    return dfs(len,0,0,1);}int main(void){    int T,n;    memset(dp,-1,sizeof(dp));    scanf("%d",&T);    int c = 1;    while(T--)    {        scanf("%d",&n);        printf("Case %d: %lld\n",c++,sum(n));    }}