HDU 5491 The Next

The Next

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 305    Accepted Submission(s): 137

 

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 $S_1\leq L\leq S_2$.
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 $(T\leq 300000)$.
Each test case consists of three integers D, S1, and S2, as described above. It is guaranteed that $0\leq D < 2^{31}$ 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

解题：直接贪心

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
inline int lowbit(int x){
    return x&-x;
}
int main(){
    int kase,cs = 1,S1,S2;
    LL D;
    scanf("%d",&kase);
    while(kase--){
        scanf("%I64d%d%d",&D,&S1,&S2);
        LL ret = D + 1,lb = lowbit(ret);
        int cnt = __builtin_popcount(ret);
        while(cnt < S1 || cnt > S2){
            int tmp = __builtin_popcount(lb - 1);
            if(S1 > cnt + tmp|| cnt > S2){
                ret += lb;
                lb = lowbit(ret);
                cnt = __builtin_popcount(ret);
            }else{
                int t = max(cnt,S1) - cnt;
                ret += (1<<t) - 1;
                break;
            }
        }
        printf("Case #%d: %I64d\n",cs++,ret);
    }
    return 0;
}