lightoj - 1032 Fast Bit Calculations （数位dp）总结

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 1or
0 (also referred to as "true" or "false"respectively). And every decimal number has a binary representation which isactually a series of bits. If a bit of a number is
1 and its next bit isalso 1 then we can say that the number has a
1 adjacent bit. Andyou 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 summationof 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[cur][s][last]表示到当前位置，前面有多少个11，前一位是last的，答案是多少

总结了一下：这个题跟上一个求多少个零一样，刚开始还没想明白，为什么要保存前面有多少个零，其实这个跟刚开始的求含有49的有多少个还是有区别的，因为前面的一个49算一次，而这个要求所有的11有多少个，那么前面含有11的，那么后面的肯定要再加上，因为后面有多少个不一样的数，就对应前面那个11多少次

一下解决了两个题（一类题），好开心

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=40;
typedef long long LL;
int N;
int dig[maxn];
LL dp[maxn][maxn][2];
LL dfs(int cur,int s,int e,int last)
{
    if(cur<0)return s;
    if(!e&&dp[cur][s][last]!=-1)return dp[cur][s][last];
    LL ans=0;
    int end=(e?dig[cur]:1);
    for(int i=0;i<=end;i++)
    {
        if(last&&i)ans+=dfs(cur-1,s+1,e&&i==end,i);
        else ans+=dfs(cur-1,s,e&&i==end,i);
    }
    if(!e)dp[cur][s][last]=ans;
    return ans;
}
LL solve(int n)
{
    int len=0;
    memset(dp,-1,sizeof(dp));
    while(n)
    {
        dig[len++]=n%2;
        n/=2;
    }
    return dfs(len-1,0,1,0);
}
int main()
{
    int T,cas=1;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&N);
        printf("Case %d: %lld\n",cas++,solve(N));
    }
    return 0;
}