hdu5491 The Next-优快云博客

本文介绍了一道算法题，要求找到给定整数D之后的下一个满足特定条件的整数Y，即Y的二进制表示中1的个数位于指定区间内。通过将D加1并检查二进制表示的1的数量来解决此问题。

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

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;
}

转载于:https://www.cnblogs.com/herumw/p/9464649.html