hdu5491 The Next

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

Total Submission(s): 740    Accepted Submission(s): 294

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

这题自己没有想到，比赛时是队友做的，后来知道了做法。先把D加1变成m，然后判断m是否满足条件，如果满足就直接输出m，如果不满足，那么有两种情况，第一种是二进制后的1的个数小于s1,那么我们只要从右往左找第一个0，使得其变为1(即加上2^i)，如果二进制后的1的个数大于s1,那么我们只要从右往左找第一个1，然后加上(2^i)，使其变为0，然后继续下去。

这种方法可行的原因是每次都是使改变最少，所以得到的值一定是成立中最小的。

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 0x7fffffff
int main()
{
    ll n,m,i,j,T,t1,t2,wei,num1=0,num;
    int a[40];
    scanf("%lld",&T);
    while(T--)
    {
        scanf("%lld%lld%lld",&n,&t1,&t2);
        n++;
        while(1){
            m=n;
            wei=0;num=0;
            while(m){
                a[++wei]=m%2;
                if(m%2==1)num++;
                m/=2;
            }
            if(num<t1){
               for(i=1;i<=wei;i++){
                    if(a[i]==0){
                        j=i;break;
                    }
               }
               n+=(1<<(j-1));
            }
            else if(num>t2){
                for(i=1;i<=wei;i++){
                    if(a[i]==1){
                        j=i;break;
                    }
                }
                n+=(1<<(j-1) );

            }
            else break;
        }
        num1++;
        printf("Case #%lld: %lld\n",num1,n);

    }
    return 0;
}