hdu 5491 The Next(ICPC合肥赛)

The Next

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 808 Accepted Submission(s): 316



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

题目大意：
就是给定一个数d,让你求二进制数中1的个数>=s1同时<=s2中的数

解题思路：
首先得到这个数二进制中1的个数，然后让这个数+1，判断当d>s2 和<s1的情况，因为找的是最小值，
所以，只需要当<s1的时候将二进制中0的个数变为1
>s1的时候，将二进制中1的个数变为0

上代码：

/**
2015 - 09 - 22 下午

Author: ITAK

Motto:

今日的我要超越昨日的我，明日的我要胜过今日的我，
以创作出更好的代码为目标，不断地超越自己。
**/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;

#define MM(a) memset(a,0,sizeof(a))

typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 10000+5;
const int mod = 1000000007;
const double eps = 1e-7;

bool arr[35];
LL t;
///得到1的个数
LL get_1(LL x)
{
    LL sum = 0;
    t = 0;
    while(x)
    {
        arr[t++] = x%2;
        if(x&1)
            sum++;
        x>>=1;
    }
    return sum;
}
int main()
{
    LL d, s1, s2, T;
    scanf("%lld",&T);
    for(LL cas=1; cas<=T; cas++)
    {
        scanf("%lld%lld%lld",&d,&s1,&s2);
        MM(arr);
        t = 0;
        d++;
        while(1)
        {
            LL ans = get_1(d);
            if(ans < s1)
            {
                for(int i=0; i<t; i++)
                    if(!arr[i])
                    {
                        d += 1<<i;
                        break;
                    }
            }
            else if(ans > s2)
            {
                for(int i=0; i<t; i++)
                    if(arr[i])
                    {
                        d += 1<<i;
                        break;
                    }
            }
            else
                break;
        }
        printf("Case #%lld: %lld\n",cas,d);
    }
    return 0;
}