CF题目集锦PART3#262div2D_html/css_WEB-ITnose

【#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<