The Next
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2080 Accepted Submission(s): 753
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
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
using namespace std;
#define ll long long
#define ms(a,b) memset(a,b,sizeof(a))
ll i,j,k,n,m;
int s[100];
ll d,s1,s2;
int main()
{
int T;
scanf("%d",&T);
int cas=1;
while(T--){
scanf("%lld%lld%lld",&n,&s1,&s2);
n++;
while(1){
m=n;
ll sum=0;
ll w=0;
while(m){
s[w++]=m&1;
if(m&1)sum++;
m>>=1;
}
if(sum<s1){
for(int i=0;i<w;i++)
if(s[i]==0){
j=i;break;
}
n+=(1<<j);
}
else if(sum>s2) {
for(int i=0;i<w;i++)
if(s[i]==1){
j=i;break;
}
n+=(1<<j);
}
else break;
}
printf("Case #%d: %lld\n",cas++,n);
}
return 0;
}
