light oj 1032(数位dp)

最新推荐文章于 2019-02-01 14:31:00 发布

转载最新推荐文章于 2019-02-01 14:31:00 发布 · 86 阅读

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原文链接：http://www.cnblogs.com/martinue/p/5547201.html

本文介绍了一道关于二进制中相邻1计数的问题——FastBitCalculations，并提供了一种解决方案，采用数位DP方法来计算从0到给定整数n的所有数中出现的连续1的总数。

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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

2147483647

Case 1: 0

Case 2: 2

Case 3: 12

Case 4: 13

Case 5: 13

Case 6: 14

Case 7: 16106127360

题意：输入n，要求从0到n的所有数的二进制数里面总共有多少“11”。

思路：将输入的数转化为二进制之后进行数位dp。dp[i][j][k]表示i位数前一位是j（由于二进制只可能是0或1）已经存在k个11的总数。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
ll dp[40][2][40];
int digit[40];
ll dfs(int pos,int pre,int st,bool limit)
{
    if(pos==0)return st;
    if(!limit&&dp[pos][pre][st]!=-1)return dp[pos][pre][st];
    ll ans=0;
    int end=limit?digit[pos]:1;
    for(int i=0; i<=end; i++)
        ans+=dfs(pos-1,i,st+(pre&&i),limit&&(i==end));
    if(!limit)
        dp[pos][pre][st]=ans;
    return ans;
}
ll get(int x)
{
    int bj=0;
    while(x)
        digit[++bj]=x%2,x/=2;
    return dfs(bj,0,0,1);
}
int main()
{
    memset(dp,-1,sizeof(dp));
    int t,o=1;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        scanf("%d",&n);
        printf("Case %d: %lld\n",o++,get(n));///light oj里面只能用lld
    }
    return 0;
}

转载于:https://www.cnblogs.com/martinue/p/5547201.html