lightoj 1032 Fast Bit Calculations 数位dp

1032 - Fast Bit Calculations

PDF (English)StatisticsForum

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

 

---

PROBLEM SETTER: MOHIUL ALAM PRINCE

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

题意：求0~n之间数位化成二进制后相邻为均为1的个数。

思路：设dp[pos][cnt]为当前考虑pos位，之前已经有cnt个11且前一位数位为1时，（pos+1）个数位与之前的数位组成的11的个数。详见代码：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN=50;typedef long long ll;int n;int bit[MAXN];ll dp[MAXN][MAXN];ll dfs(int pos,int pre,int cnt,int flag){    if(pos == -1) return cnt;    if(flag && dp[pos][cnt]!=-1 && pre) return dp[pos][cnt];    int x=flag ? 1: bit[pos];    ll ans=0;    for(int i=0;i<=x;i++){        ans+=dfs(pos-1,i==1,cnt+(i==1 && pre),flag || i<x);    }    if(flag &&pre) dp[pos][cnt]=ans;    return ans;}ll solve(int x){    int len=0;    while(x)    {        bit[len++]=x%2;        x/=2;    }    return dfs(len-1,0,0,0);}int main(){    //freopen("text.txt","r",stdin);    int T,kase=0;    scanf("%d",&T);    memset(dp,-1,sizeof(dp));    while(T--){        kase++;        scanf("%d",&n);        printf("Case %d: %lld\n",kase,solve(n));    }    return 0;}