POJ 3252 Round Numbers （数位dp）

题意：

求区间内  二进制  0 的个数大于等于 1  的 个数。

思路：

显然数位dp。

但要考虑二进制 0 和1 的个数。不太好做。

我们平常写数位dp 都是十进制的。

如果我们直接按二进制直接数位dp  就和十进制一样了。。

令dp[i][j][k] 表示 第i位，  目前有j 个0  k 个1的方案数。

然后就是很基本的数位dp了。

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 35;

int bit[maxn];
int dp[maxn][maxn][maxn];

int dfs(int len, int num0, int num1, bool lead, bool limit){
    if (len < 0){
//        if (lead) return 0;
        if (num0 >= num1){
            return 1;
        }
        else {
            return 0;
        }
    }
    if (!lead && !limit && dp[len][num0][num1] != -1){
        return dp[len][num0][num1];
    }

    int up = limit ? bit[len] : 1;
    int ans = 0;
    for (int i = 0; i <= up; ++i){
        if (lead){
            if (i == 0)
                ans += dfs(len-1, num0, num1, true, limit && i == up);
            else
                ans += dfs(len-1, num0, num1 + 1, false, limit && i == up);
        }
        else {
            if (i == 0){
                ans += dfs(len-1, num0 + 1, num1, false, limit && i == up);
            }
            else {
                ans += dfs(len-1, num0, num1+1, false, limit && i == up);
            }
        }
    }

    if (!lead && !limit){
        dp[len][num0][num1] = ans;
    }
    return ans;
}

int solve(int x){
    int len = 0;
    while(x){
        bit[len++] = x & 1;
        x >>= 1;
    }

    memset(dp,-1,sizeof dp);
    return dfs(len-1, 0, 0, true, true);
}
int main(){
    int a, b;
    scanf("%d %d",&a, &b);
    printf("%d\n", solve(b) - solve(a-1));
    return 0;
}

Round Numbers Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 14198   Accepted: 5663 Description 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 Source USACO 2006 November Silver

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