The Next
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 808 Accepted Submission(s): 316
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.
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.
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
题目大意：
就是给定一个数d,让你求二进制数中1的个数>=s1同时<=s2中的数
解题思路：
首先得到这个数二进制中1的个数，然后让这个数+1，判断当d>s2 和<s1的情况，因为找的是最小值，
所以，只需要当<s1的时候将二进制中0的个数变为1
>s1的时候，将二进制中1的个数变为0
上代码：
/**
2015 - 09 - 22 下午
Author: ITAK
Motto:
今日的我要超越昨日的我，明日的我要胜过今日的我，
以创作出更好的代码为目标，不断地超越自己。
**/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;
#define MM(a) memset(a,0,sizeof(a))
typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 10000+5;
const int mod = 1000000007;
const double eps = 1e-7;
bool arr[35];
LL t;
///得到1的个数
LL get_1(LL x)
{
LL sum = 0;
t = 0;
while(x)
{
arr[t++] = x%2;
if(x&1)
sum++;
x>>=1;
}
return sum;
}
int main()
{
LL d, s1, s2, T;
scanf("%lld",&T);
for(LL cas=1; cas<=T; cas++)
{
scanf("%lld%lld%lld",&d,&s1,&s2);
MM(arr);
t = 0;
d++;
while(1)
{
LL ans = get_1(d);
if(ans < s1)
{
for(int i=0; i<t; i++)
if(!arr[i])
{
d += 1<<i;
break;
}
}
else if(ans > s2)
{
for(int i=0; i<t; i++)
if(arr[i])
{
d += 1<<i;
break;
}
}
else
break;
}
printf("Case #%lld: %lld\n",cas,d);
}
return 0;
}
