Description
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.
The first line contains three space-separated integers l, r, k (1 ≤ l ≤ r ≤ 1012; 1 ≤ k ≤ min(106, r - l + 1)).
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.
8 15 3
1
2
10 11
8 30 7
0
5
14 9 28 11 16
Operation 
represents the operation of bitwise exclusive OR. In other words, it is the XOR operation.
正解：分类讨论
解题报告：
今天考试原题，直接上题解：
分类讨论。
首先注意到(2x) ⊕ (2x + 1) = 1。一个直接推论就是(2x) ⊕ (2x + 1) ⊕ (2x + 2) ⊕(2x + 3) = 0。而k ≥ 5的时候一定可以选出四个数其异或和为0(例如,如果l是偶数,那么选l, l + 1, l + 2, l + 3,否则选l + 1, l + 2, l + 3, l + 4。这样总是合法的,因为r ≥ l + 4)。
然后按照k的值分类讨论。
k = 1。答案就是l。
k = 2。如果r = l + 1,那么答案是min(l ⊕ r, l),其中l表示只选一个数,l ⊕ r表示选两个数。否则答案一定为1。当r > l + 1的时候,如果l是偶数,那么答案是l ⊕ (l + 1),否则答案是(l + 1) ⊕ (l + 2)。
k = 3。答案不会超过1,因为从[l, r]中选两个数一定可以使得答案为1。我们关心的是三个数的异或和为0的情况。这3个数的最高位不可能相同(否则异或起来的最高位为1)。设这三个数为x, y, z(x < y <z),且y = 2 ^k + b, z = 2^ k + c, b < c。那么x ⊕ b ⊕ c = 0。为了让l ≤ x, y, z ≤ r,我们需要使得x尽量大而c尽量小。假设x ≥ 2 ^(k−1) ,那么可以推出z ≥ 2^( k−1) + 2^ k 。我们需要使x尽量大而c尽量小,那么令x = 2 ^(k − 1), z = 2 ^(k−1) + 2^ k ,可以得到y = z − 1。满足要求。那么如果x < 2^( k−1) 呢?我们会发现:此时z没必要≥ 2^ k ,因为取y = 2^( k−1 + b), z = 2 ^(k−1 )+ c依然满足条件。
所以枚举k并检查x = 2 ^k − 1, z = 2^ k + 2^( k−1) , y = z − 1这三个数是否在[l, r]内,如果在的话,就找到了一组解。
k = 4。如果r − l ≥ 4,那么答案一定为0。否则,枚举{x|x ∈ Z ∧ l ≤ x ≤ r}的所有子集,求异或和的最小值。
我的代码后面分类讨论部分写丑了。
1 //It is made by jump~ 2 #include <iostream> 3 #include <cstdlib> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 #include <algorithm> 8 #include <ctime> 9 #include <vector> 10 #include <queue> 11 #include <map> 12 #include <set> 13 using namespace std; 14 typedef long long LL; 15 LL l,r,len; 16 int k; 17 LL jilu1,jilu2,jilu3,num; 18 19 inline int getint() 20 { 21 int w=0,q=0; 22 char c=getchar(); 23 while((c<'0' || c>'9') && c!='-') c=getchar(); 24 if (c=='-') q=1, c=getchar(); 25 while (c>='0' && c<='9') w=w*10+c-'0', c=getchar(); 26 return q ? -w : w; 27 } 28 29 inline LL getlong() 30 { 31 LL w=0,q=0; 32 char c=getchar(); 33 while((c<'0' || c>'9') && c!='-') c=getchar(); 34 if (c=='-') q=1, c=getchar(); 35 while (c>='0' && c<='9') w=w*10+c-'0', c=getchar(); 36 return q ? -w : w; 37 } 38 39 inline LL check(){ 40 LL t=0; 41 while( ((1LL<<t)-1LL) < l) t++; 42 if(( (1LL<<t)+(1LL<<(t-1LL)) )<=r) return t; 43 return -1; 44 } 45 46 inline void work(){ 47 LL now; 48 l=getlong(); r=getlong(); k=getint(); len=r-l+1; 49 if(k==1) printf("%I64d 1 %I64d",l,l); 50 else if(k==2) { 51 if(l&1) { 52 if(r-l+1==2){ 53 if((l^r)<l) printf("%I64d 2 %I64d %I64d",l^r,l,r); 54 else printf("%I64d 1 %I64d",l,l); 55 } 56 else printf("1 2 %I64d %I64d",l+1,l+2); 57 } 58 else printf("1 2 %I64d %I64d",l,l+1); 59 } 60 else if(k==3) { 61 if((now=check())!=-1) { 62 printf("0 3 "); printf("%I64d ",(1LL<<now)-1LL); 63 printf("%I64d %I64d",(1LL<<now)+(1LL<<(now-1LL))-1LL, (1LL<<now)+(1LL<<(now-1LL) ) ); 64 } 65 else { printf("1 2 "); if(l&1) printf("%I64d %I64d",l+1,l+2); else printf("%I64d %I64d",l,l+1); } 66 } 67 else{ 68 if(l&1) { 69 if(r-l+1>4) printf("0 4 %I64d %I64d %I64d %I64d",l+1,l+2,l+3,l+4); 70 else{ 71 now=l; num=1; for(int i=0;i<4;i++) for(int j=i+1;j<4;j++) if( ((l+i)^(l+j)) < now) now=((l+i)^(l+j)),num=2,jilu1=l+i,jilu2=l+j; 72 if( (l^(l+1)^(l+2) ) < now) now=(l^(l+1)^(l+2) ),num=3,jilu1=l,jilu2=l+1,jilu3=l+2; if( (l^(l+1)^(l+3) ) < now) now=(l^(l+1)^(l+3) ),num=3,jilu1=l,jilu2=l+1,jilu3=l+3; 73 if( (l^(l+2)^(l+3) ) < now) now=(l^(l+2)^(l+3) ),num=3,jilu1=l,jilu2=l+2,jilu3=l+3; if( ((l+1)^(l+2)^(l+3) ) < now) now=((l+1)^(l+2)^(l+3)),num=3,jilu1=l+1,jilu2=l+2,jilu3=l+3; 74 if( (l^(l+1)^(l+2)^(l+3) ) < now) now=(l^(l+1)^(l+2)^(l+3) ),num=4; 75 printf("%I64d %I64d ",now,num); if(num==1) printf("%I64d",now); else if(num==2) printf("%I64d %I64d",jilu1,jilu2); else if(num==3) printf("%I64d %I64d %I64d",jilu1,jilu2,jilu3); else printf("%I64d %I64d %I64d %I64d",l,l+1,l+2,l+3); 76 } 77 } 78 else printf("0 4 %I64d %I64d %I64d %I64d",l,l+1,l+2,l+3); 79 } 80 } 81 82 int main() 83 { 84 work(); 85 return 0; 86 }