hdu 5491 The Next

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

3 11 2 4 22 3 3 15 2 5

 

 

Sample Output

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

 

 

Source

2015 ACM/ICPC Asia Regional Hefei Online

 暴力枚举  不多说

#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int oo = 1e9;
const double PI = acos(-1);
const int N = 1e3;
int binary[N], one, k;
void get(LL n)
{
    one = k = 0;
    while(n)
    {
        if(n % 2 == 1)
        {
            binary[k] = 1;
            one++;
        }
        else binary[k] = 0;
        k++;
        n /= 2;
    }
    k--;
}
int main()
{
    int T, s1, s2, L, cas=1;
    LL ans;
    scanf("%d", &T);
    while(T--)
    {
        memset(binary, 0, sizeof(binary));
        scanf("%d %d %d", &L, &s1, &s2);
        ans = L;
        do
        {
            ans ++;
            get(ans);
            if(one < s1)
            {
                int j = s1 - one, id=0;
                while(j)
                {
                    if(binary[id] == 0)
                    {
                        binary[id] = 1;
                        j--;
                    }
                    id++;
                }
                break;
            }

        }while(one < s1 || one > s2);
        ans = 0;
        
        for(int i = k; i >= 0; i--)
            ans = ans * 2 + binary[i];

        printf("Case #%d: %lld\n", cas++, ans);
    }
    return 0;
}

　　

 

转载于:https://www.cnblogs.com/PersistFaith/p/4844424.html