HDU5491-The Next

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本文介绍了一道算法题目，任务是找到大于给定整数的下一个整数，该整数的二进制表示中1的个数落在指定范围内。文章提供了详细的解题思路和完整的C++代码实现。

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The Next

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2088 Accepted Submission(s): 758

Problem Description

Let

L denote the number of 1s in integer

D ’s binary representation. Given two integers

S1 and

S2 , we call

D a WYH number if

S1≤L≤S2 .
With a given

D , we would like to find the next WYH number

Y , which is JUST larger than

D . In other words,

Y is the smallest WYH number among the numbers larger than

D . Please write a program to solve this problem.

Input

The first line of input contains a number

T indicating the number of test cases (

T≤300000 ).
Each test case consists of three integers

D ,

S1 , and

S2 , as described above. It is guaranteed that

0≤D<231 and

D is a WYH number.

Output

For each test case, output a single line consisting of “Case #X: Y”.

X is the test case number starting from 1.

Y is the next WYH number.

Sample Input

Sample Output

  
   Case #1: 12
Case #2: 25
Case #3: 17

Source

2015 ACM/ICPC Asia Regional Hefei Online

Recommend

wange2014

题意：给你一个数，并且它二进制中1的个数是大于等于s1，小于等于s2，让你找出最小的比这个数大的并且满足二进制中1的个数大于等于s1，小于等于s2

解题思路：分两种情况，1的个数大于s2，则从低位开始找到第一个1加1，然后向前进位，1的个数小于s1，则从低位开始找到第一个0，然后变一

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

char ch[105];
LL n, m;
int s1, s2;

int main()
{
	int t,cas=0;
	scanf("%d", &t);
	while (t--)
	{
		scanf("%lld%d%d", &n, &s1, &s2);
		memset(ch, '0', sizeof ch);
		n++;
		int sum = 0, cnt = 0;
		while (n)
		{
			ch[cnt++] = n % 2+'0';
			n /= 2;
			if (ch[cnt - 1] == '1') sum++;
		}
		while (1)
		{
			if (sum >= s1&&sum <= s2) break;
			if (sum < s1)
			{
				for (int i = 0; i < cnt; i++)
					if (ch[i] == '0') { sum++; ch[i] = '1'; break; }
			}
			else
			{
				int k = 0,cnt;
				while (ch[k] == '0') k++;
				ch[k++] = '0', cnt = 1,sum--;
				while (ch[k] + cnt == '2')
				{
					ch[k++] = '0';
					sum--;
				}
				ch[k] = '1';
				sum++;
			}
		}
		m = 0;
		for (int i = 0; i < 35; i++)
			if (ch[i] == '1') m |= (1LL << i);
		printf("Case #%d: %lld\n",++cas, m);
	}
	return 0;
}