hdu 5491 The Next（暴力枚举）

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≤).
Each test case consists of three integers D, S1, and S2, as described above. It is guaranteed that ≤D< 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 . Y is the next WYH number.

Sample Input

 

Sample Output

Case #:
Case #:
Case #: 

 

Source

2015 ACM/ICPC Asia Regional Hefei Online

 

不想写，找了个别人的.

思路：

考虑通过比较当前的1的数目来进行最小的数的修正。

先将D加1，然后计算出D的1的数目tot，这时候比较tot和s1,s2的大小，这时候有两种情况：

1、tot<s1，这时我需要增加1的数目，因为要最小，所以我从右开始找到一个0（位置假设为i），将D加上2^i。

2、tot>s2，这时我需要减少1的数目，所以我从右开始找到一个1，将D加上2^i。

如此循环上述过程，直到符合tot符合s1,s2的条件。

上面操作的原因是：我加上2^i，就是将那一位由1变为0或者由0变为1，而我是从右边开始寻找的，可以保证每次所做的改变都是最小的。同时我的D是一直增加的，所以当条件满足时，就是那个最小的。

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<math.h>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<map>
 #include<vector>
 #include<stdlib.h>
 #include <stack>
 using namespace std;
 int dirx[]={,,-,};
 int diry[]={-,,,};
 #define PI acos(-1.0)
 #define max(a,b) (a) > (b) ? (a) : (b)
 #define min(a,b) (a) < (b) ? (a) : (b)
 #define ll long long
 #define eps 1e-10
 #define MOD 1000000007
 #define N 100
 #define inf 1e12
 ll d,s1,s2;
 ll num[N];
 ll k;
 ll count_(ll dd){
    memset(num,,sizeof(num));
    ll cnt=;
    k=;
        while(dd){
            if(dd&) cnt++;
            num[k++]=(dd&);
            dd>>=;
        }
        num[k++]=;

    return cnt;
 }
 int main()
 {
    ll ac=;
    ll t;
    scanf("%I64d",&t);
    while(t--){
        scanf("%I64d%I64d%I64d",&d,&s1,&s2);

        ll dd=d+;

        while(){
            ll tmp=count_(dd);

            if(tmp<s1){
               for(ll i=;i<k;i++){
                   if(num[i]==){
                      dd+=(<<i);
                      break;
                   }
               }
            }else if(tmp>s2){
               for(ll i=;i<k;i++){
                   if(num[i]==){
                      dd+=(<<i);
                      break;
                   }
               }
            }else{
               break;
            }
        }
        printf("Case #%I64d: ",++ac);
        printf("%I64d\n",dd);

    }
     return ;
 }