POJ 3252 Round Numbers 组合数学

最新推荐文章于 2020-02-11 17:56:01 发布

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原文链接：http://www.cnblogs.com/joeylee97/p/6978725.html

解决奶牛们通过二进制数字游戏决定先后顺序的问题，利用组合数学找出特定范围内圆整数的数量，即二进制表示中零的数量不小于一的数量的正整数。

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 13381		Accepted: 5208

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

Source

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<deque>
#include<iomanip>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN 18
#define N 33
#define MOD 1000000
#define INF 1000000009
const double eps = 1e-9;
const double PI = acos(-1.0);
/*
组合数学 找规律
如果二进制表示的长度为len（第一位必须为1）
那么要求在len-1中挑选 大于(len-1)/2个0
求[L,R] 里面有多少roundnumber 前缀和的思想
小于R的减去小于L的
*/
int C[N][N] = { 0 };
int bits[N];
void Init()//打组合数表
{
    for (int i = 0; i <= N; i++)
    {
        for (int j = 0; j <= i; j++)
        {
            if (j == 0 || j == i)
                C[i][j] = 1;
            else
                C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
        }
    }
    C[0][0] = 1;
    return;
}
int Count(int x)//小于x的Round有多少
{
    if (x <= 1)
        return 0;
    int len = 0;
    while (x)
    {
        if (x & 1)bits[len++] = 1;
        else bits[len++] = 0;
        x /= 2;
    }
    int ans = 0, cnt0 = 0, cnt1 = 0;
    for (int i = 0; i < len; i++)
    {
        if (bits[i] == 1)
            cnt1++;
        else
            cnt0++;
    }
    if (cnt0 >= cnt1) ans++;//本身是不是一个Round number
    for (int i = len - 1; i > 0; i--)//位数小于它的，有多少Round Number
    {
        if (i % 2 == 0)ans += (1 << (i - 1)) / 2;
        else ans += ( (1 << (i - 1)) - C[i - 1][(i - 1) / 2]) / 2;
    }
    cnt0 = 0, cnt1 = 1;
    for (int i = len - 2; i >= 0; i--)
    {
        if (bits[i] == 1)
        {
            for (int j = i; j >= 0 && j + cnt0 + 1 >= i - j + cnt1; j--)
                ans += C[i][j];
            cnt1++;
        }
        else
            cnt0++;
    }
    return ans;
}
int main()
{
    Init();
    int l, r;
    while (scanf("%d%d", &l, &r) != EOF)
    {
        printf("%d\n", Count(r) - Count(l-1));
    }
}

转载于:https://www.cnblogs.com/joeylee97/p/6978725.html