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时间:2021-07-01 10:21:17 帮助过:29人阅读

【#262 div 2 D. Little Victor and Set】


【原题】

D. Little Victor and Set

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Victor adores the sets theory. Let us remind you that a set is a group of numbers where all numbers are pairwise distinct. Today Victor wants to find a set of integers S that has the following properties:

  • for all x the following inequality holds l?≤?x?≤?r;
  • 1?≤?|S|?≤?k;
  • lets denote the i-th element of the set S as si; value must be as small as possible.
  • Help Victor find the described set.

    Input

    The first line contains three space-separated integers l,?r,?k (1?≤?l?≤?r?≤?1012; 1?≤?k?≤?min(106,?r?-?l?+?1)).

    Output

    Print the minimum possible value of f(S). Then print the cardinality of set |S|. Then print the elements of the set in any order.

    If there are multiple optimal sets, you can print any of them.

    Sample test(s)

    input

    8 15 3

    output

    1210 11

    input

    8 30 7

    output

    0514 9 28 11 16

    Note

    Operation represents the operation of bitwise exclusive OR. In other words, it is the XOR operation.


    【题意】给定范围L和R,在这之间选P个不同的自然数,其中1<=P<=k,求选出的数最小异或和及某个方案。

    【分析】很显然的结论,K^(K+1)=1,其中K是偶数。当K>3时,我们可以选连续的4个自然数使异或和为0。(当然注意要特判R-L+1的大小)。当K=1时,就是L。当K=2时,显然只能构造异或为1的情况。

    所有的推论都指向一个问题:当K=3的一般情况怎么做?

    【题解】对于那个情况,我一直觉得能贪心构造,但是怎么也想不出简单易行且效率高的算法。

    其实很简单。我们设L<=X

    在二进制中,异或和为0的情况是1,1,0或0,0,0。显然Z的第一位是1,然后X和Y是0。

    因为是贪心,我们要尽量使Y靠近Z(因为如果Z符合范围,Y显然越大越好)。

    那么第二位我们就让Y靠近Z。我们把Z那位设成0,X和Y都设成1,即如下形式:

    110000000

    101111111

    011111111

    当然脑补可能会萎...

    为了少特判,我在R-L+1小的时候直接暴力寻找。

    【代码】

    #include#include#include#define E endl#define INF 999999999999999ll#define RE return 0using namespace std;typedef long long LL;LL len,sum,ans,C,wri[15],temp[15],i,S,L,R,k,x,z;inline void DFS(LL now,LL C,LL sum){  if (now==R+1)   {    if (sum>=ans||!C) return;len=C;ans=sum;    for (int i=1;i<=C;i++) wri[i]=temp[i];    return;  }  if (now>R) return;  DFS(now+1,C,sum);if (C+1>k) return;  temp[C+1]=now;DFS(now+1,C+1,sum^now);}int main(){  cin>>L>>R>>k;  if (L==R) {cout<=L) {cout<<0<

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