【Codeforces Round 262 (Div 2)D】【构造】Little Victor and Set 集合最多取k数使得异或值尽可能小
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.
Examples
input
8 15 3
output
1 2 10 11
input
8 30 7
output
0 5 14 9 28 11 16
Note
Operation 
represents the operation of bitwise exclusive OR. In other words, it is the XOR operation.
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
LL l, r, k, top;
LL a[20], b[20];
LL ans, ansnum;
void dfs(LL p, LL v, LL num)
{
if (num && v < ans)
{
ans = v;
ansnum = num;
MC(b, a);
}
if (p > top || num == k)return;
dfs(p + 1, v, num);
a[num+1] = p;
dfs(p + 1, v^p, num + 1);
}
int main()
{
while (~scanf("%lld%lld%lld", &l,&r,&k))
{
if (k == 3)
{
bool flag = 0;
for (LL i = 2; i <= r; i <<= 1)
{
LL half = i / 2;
LL z = i - 1;
LL y = z + half;
LL x = y^z;
if (z >= l&&x <= r)
{
puts("0");
puts("3");
printf("%lld %lld %lld\n", x, y, z);
flag = 1;
break;
}
}
if (flag)continue;
}
top = min(r, l + 4);
ans = 1e18;
dfs(l, 0, 0);
printf("%lld\n", ans);
printf("%lld\n", ansnum);
for (int i = 1; i <= ansnum; ++i)
{
printf("%lld ", b[i]);
}puts("");
}
return 0;
}
/*
【题意】
给你一个区间[l,r]，区间内最少取1个数，最多取k（1e12）个数。1<=k<=min(1e6,r-l+1)。
我们相使得所取数的异或值尽可能小。
【类型】
暴力
【分析】
首先，连续取2个相邻数的话，异或和很可能就为1
连续取4个相邻数的话，异或和很可能就为0
显然的是，在5个连续数a,b,c,d,e中，
a^b^c^d和b^c^d^e必然有一个为0。
于是我们可以在小范围内，暴力dfs出一个解。
即k>=4和k为1或2的时候，我们都有决策了。
于是，剩下需要注意的就是k==3的时候。
关键就变成——如何取3个数并使得异或值为0。
然后我们可以采取的贪心策略是——
首先，如果最大数和最小数的位数间隙为2或者更大的话，我们就可以有下面这种构造方案：
1100
1000
0100
于是，剩下最大数和最小数的位数间隙为1。
我们可以用以下
1100000……
1011111……
0111111……
的方式做构造，这样3个数的范围会尽可能缩小。而且是上下界同时缩小，就可以了。
诗诗的做法是——
取l最高位的1
11..0..0
10..1..0
l: 1..1..0
然后对于l，如果这一位为1，那么就走01；如果为0，就走00，
显然不影响三个数的大小关系且使得a>b>c中a的值尽可能小
其实本质相同。
*/