Round Numbers(数位dp)

The cows, as you know, have no fingers or thumbs and thus are unable to play Scissors, Paper, Stone' (also known as 'Rock, Paper, Scissors', 'Ro, Sham, Bo', and a host of other names) in order to make arbitrary decisions such as who gets to be milked first. They can't even flip a coin because it's so hard to toss using hooves.

They have thus resorted to "round number" matching. The first cow picks an integer less than two billion. The second cow does the same. If the numbers are both "round numbers", the first cow wins,
otherwise the second cow wins.

A positive integer N is said to be a "round number" if the binary representation of N has as many or more zeroes than it has ones. For example, the integer 9, when written in binary form, is 1001. 1001 has two zeroes and two ones; thus, 9 is a round number. The integer 26 is 11010 in binary; since it has two zeroes and three ones, it is not a round number.

Obviously, it takes cows a while to convert numbers to binary, so the winner takes a while to determine. Bessie wants to cheat and thinks she can do that if she knows how many "round numbers" are in a given range.

Help her by writing a program that tells how many round numbers appear in the inclusive range given by the input (1 ≤ Start < Finish ≤ 2,000,000,000).

Input

Line 1: Two space-separated integers, respectively Start and Finish.

Output

Line 1: A single integer that is the count of round numbers in the inclusive range Start.. Finish

Sample Input

2 12

Sample Output

6

题意:求[l, r]中满足二进制0的个数大于1的个数,的数目

题解:简单数位dp,直接看代码

//#include"bits/stdc++.h"
//#include
//#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define LL long long
#define ULL unsigned long long
#define MT(a,b) memset(a,b,sizeof(a))
#define lson l, mid, node << 1
#define rson mid + 1, r, node << 1 | 1
const int INF  =  0x3f3f3f3f;
const int O    =  1e6;
const int mod  =  10007;
const int maxn =  1e4+5;
const double PI  =  acos(-1.0);
const double E   =  2.718281828459;

const int _ = 1;
const int __ = 2;
const int ___ = 3;

int dp[35][35][35]; // dp[pos][num0][num1]表示pos为0和1的个数已经为num0和num1,满足条件的个数
int a[35];

int dfs(int pos, int num0, int num1, int lead, int flag){
    if(pos == -1) return num0 >= num1;
    if(!flag && !lead && dp[pos][num0][num1] != -1) return dp[pos][num0][num1];
    
    int sum = 0;
    int up =  flag ? a[pos] : 1;
    
    for(int i=0; i<=up; i++){
        sum += dfs(pos-1, (!i && lead) ? 0 : num0 + !i, (!i && lead) ? 0 : num1 + i, !i && lead, flag && i==a[pos]);
//        或者:
//        if(i == 0 && lead) sum += dfs(pos -1, 0, 0, lead && i == 0, flag && i == a[pos]);
//        else if(i !=0 && lead) sum += dfs(pos -1, 0, 1, lead && i == 0, flag && i == a[pos]);
//        else if(i == 0 && !lead) sum += dfs(pos-1, num0+1, num1, lead && i == 0, flag && i == a[pos]);
//        else if(i !=0 && !lead) sum += dfs(pos-1, num0, num1 +1, lead && i == 0, flag && i == a[pos]);
    }
    
    if(!flag && !lead) dp[pos][num0][num1] = sum;
    return sum;
}

int solve(int x){
    int cnt = 0;
    while(x) { a[cnt ++] = x & 1; x >>= 1; }
    return dfs(cnt -1, 0, 0, 1, 1);
}

int main(){
    int l, r; scanf("%d%d", &l, &r);
    MT(dp, -1);
    printf("%d\n", solve(r) - solve(l-1));
    return 0;
}

 

你可能感兴趣的:(数位dp)