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本文介绍了一种算法，用于找到大于给定整数D的最小WYH数。通过输入三个整数D、S1和S2，读者可以了解如何确定满足条件的下个WYH数。

The Next

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 943 Accepted Submission(s): 389

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

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define ll long long
using namespace std;
int num[40];
int sum[40];
ll work(){
    ll a,l,r;
    scanf("%lld%lld%lld",&a,&l,&r);
    memset(num,0,sizeof(num));
    memset(sum,0,sizeof(sum));
    int cnt = 0;
    while(a){
        num[cnt++] = a&1;
        a >>= 1;
    }
    for(int i = 33;i >= 0; i--){
        sum[i] = sum[i+1]+num[i];
    }
    for(int i = 0;i < 34; i++){
        if(num[i] == 0){
            int t = 1+sum[i];
            if(t+i>=l&&t<=r){
                int need = max(l-t,0ll);
                num[i] = 1;
                for(int j = 0;j < i; j++)
                    num[j] = 0;
                for(int j = 0;j < need; j++)
                    num[j] = 1;
                break;
            }
        }
    }
    ll ans = 0;
    for(int i = 0;i < 34; i++)
        if(num[i])
            ans += (1ll<<i);
    return ans;
}

int main(){
    int t,tt=1;
    scanf("%d",&t);
    while(t--){
        ll ans = work();
        printf("Case #%d: %I64d\n",tt++,ans);
    }
    return 0;
}