Codeforces Round #262 (Div. 2) 1004
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
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.
【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<Y<Z<=R，然后来贪心构造他们。
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
wa了很多次，
1.没有用long long
2.只有l^(l+1) l为偶数时，才能异或值为1
3.当k>=4但是不存在4个数异或为0的时候，没考虑3个也可能为0
4.1<<35超过int 得写成（long long）1<<35
5.当2个异或不是1时，应该判断他的值和l的大小
#include <cstring>
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <map>
#include <cstdlib>
#define M(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
using namespace std;
long long l,r,k;
int main()
{
scanf("%I64d%I64d%I64d",&l,&r,&k);
if(k==) printf("%I64d\n1\n%I64d\n",l,l);
else if(k==)
{
if(l%==)
printf("1\n2\n%I64d %I64d\n",l,l+);
else if(l+<=r)
printf("1\n2\n%I64d %I64d\n",l+,l+);
else if(((l)^(l+))<l) {
//cout<<(((l)^(l+1))-l)<<endl;
printf("%I64d\n2\n%I64d %I64d\n",(l)^(l+),l,l+);
}
else printf("%I64d\n1\n%I64d\n",l,l);
}
else if(k>=)
{
if(l%==)
printf("0\n4\n%I64d %I64d %I64d %I64d\n",l,l+,l+,l+);
else if(l+<r)
printf("0\n4\n%I64d %I64d %I64d %I64d\n",l+,l+,l+,l+);
else if(((l)^(l+)^(l+)^(l+))==)printf("0\n4\n%I64d %I64d %I64d %I64d\n",l,l+,l+,l+);
else
{
int count1 = ;
long long tem1 = r;
while(tem1>)
{
tem1 = tem1>>;
count1++;
}
//cout<<count1<<endl;
int cnt = ;
long long ans1 = ;
long long ans2 = ;
for(int i = count1-;i>=;i--)
{
if(((r>>i)&)==)
{
if(cnt == )
{
ans1 = ans1|((long long)<<i);
cnt++;
}
else if(cnt >= )
{
ans2 = ans2|((long long)<<i);
cnt++;
}
}
else
{
if(cnt>)
{
ans1 = ans1|((long long)<<i);
ans2 = ans2|((long long)<<i);
}
}
}
if(ans2<l)
{
if(l%==)
printf("1\n2\n%I64d %I64d\n",l,l+);
else printf("1\n2\n%I64d %I64d\n",l+,l+);
}
else printf("0\n3\n%I64d %I64d %I64d\n",ans1,ans2,r);
}
}
else
{
int count1 = ;
long long tem1 = r;
while(tem1>)
{
tem1 = tem1>>;
count1++;
}
//cout<<count1<<endl;
int cnt = ;
long long ans1 = ;
long long ans2 = ;
for(int i = count1-;i>=;i--)
{
if(((r>>i)&)==)
{
if(cnt == )
{
ans1 = ans1|((long long)<<i);
cnt++;
}
else if(cnt >= )
{
ans2 = ans2|((long long)<<i);
cnt++;
}
}
else
{
if(cnt>)
{
ans1 = ans1|((long long)<<i);
ans2 = ans2|((long long)<<i);
}
}
//cout<<ans2<<' '<<ans1<<endl;
}
//cout<<ans2<<' '<<ans1<<endl;
if(ans2<l)
{
if(l%==)
printf("1\n2\n%I64d %I64d\n",l,l+);
else printf("1\n2\n%I64d %I64d\n",l+,l+);
}
else printf("0\n3\n%I64d %I64d %I64d\n",ans1,ans2,r);
}
return ;
}