Codeforces Round #262 (Div. 2) 1004

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

  1 #include <cstring>
  2 
  3 #include <iostream>
  4 
  5 #include <algorithm>
  6 
  7 #include <cstdio>
  8 
  9 #include <cmath>
 10 
 11 #include <map>
 12 
 13 #include <cstdlib>
 14 
 15 #define M(a,b) memset(a,b,sizeof(a))
 16 
 17 #define INF 0x3f3f3f3f
 18 
 19 using namespace std;
 20 
 21 
 22 
 23 long long l,r,k;
 24 
 25 
 26 
 27 int main()
 28 
 29 {
 30 
 31     scanf("%I64d%I64d%I64d",&l,&r,&k);
 32 
 33     if(k==1) printf("%I64d\n1\n%I64d\n",l,l);
 34 
 35     else if(k==2)
 36 
 37     {
 38 
 39         if(l%2==0)
 40 
 41         printf("1\n2\n%I64d %I64d\n",l,l+1);
 42 
 43         else if(l+2<=r)
 44 
 45         printf("1\n2\n%I64d %I64d\n",l+1,l+2);
 46 
 47         else if(((l)^(l+1))<l) {
 48 
 49                 //cout<<(((l)^(l+1))-l)<<endl;
 50 
 51                 printf("%I64d\n2\n%I64d %I64d\n",(l)^(l+1),l,l+1);
 52 
 53         }
 54 
 55         else printf("%I64d\n1\n%I64d\n",l,l);
 56 
 57     }
 58 
 59     else if(k>=4)
 60 
 61     {
 62 
 63         if(l%2==0)
 64 
 65           printf("0\n4\n%I64d %I64d %I64d %I64d\n",l,l+1,l+2,l+3);
 66 
 67         else if(l+3<r)
 68 
 69           printf("0\n4\n%I64d %I64d %I64d %I64d\n",l+1,l+2,l+3,l+4);
 70 
 71         else if(((l)^(l+1)^(l+2)^(l+3))==0)printf("0\n4\n%I64d %I64d %I64d %I64d\n",l,l+1,l+2,l+3);
 72 
 73          else
 74 
 75     {
 76 
 77         int count1 = 0;
 78 
 79         long long tem1 = r;
 80 
 81         while(tem1>0)
 82 
 83         {
 84 
 85             tem1 = tem1>>1;
 86 
 87             count1++;
 88 
 89         }
 90 
 91         //cout<<count1<<endl;
 92 
 93         int cnt = 0;
 94 
 95         long long ans1 = 0;
 96 
 97         long long ans2 = 0;
 98 
 99         for(int i = count1-1;i>=0;i--)
100 
101         {
102 
103             if(((r>>i)&1)==1)
104 
105             {
106 
107                 if(cnt == 0)
108 
109                 {
110 
111                   ans1 = ans1|((long long)1<<i);
112 
113                   cnt++;
114 
115                 }
116 
117                 else if(cnt >= 1)
118 
119                 {
120 
121                     ans2 = ans2|((long long)1<<i);
122 
123                     cnt++;
124 
125                 }
126 
127             }
128 
129             else
130 
131             {
132 
133                 if(cnt>1)
134 
135                 {
136 
137                     ans1 = ans1|((long long)1<<i);
138 
139                     ans2 = ans2|((long long)1<<i);
140 
141                 }
142 
143             }
144 
145         }
146 
147         if(ans2<l)
148 
149         {
150 
151             if(l%2==0)
152 
153                 printf("1\n2\n%I64d %I64d\n",l,l+1);
154 
155             else printf("1\n2\n%I64d %I64d\n",l+1,l+2);
156 
157         }
158 
159         else printf("0\n3\n%I64d %I64d %I64d\n",ans1,ans2,r);
160 
161     }
162 
163     }
164 
165     else
166 
167     {
168 
169         int count1 = 0;
170 
171         long long tem1 = r;
172 
173         while(tem1>0)
174 
175         {
176 
177             tem1 = tem1>>1;
178 
179             count1++;
180 
181         }
182 
183         //cout<<count1<<endl;
184 
185         int cnt = 0;
186 
187         long long ans1 = 0;
188 
189         long long ans2 = 0;
190 
191         for(int i = count1-1;i>=0;i--)
192 
193         {
194 
195             if(((r>>i)&1)==1)
196 
197             {
198 
199                 if(cnt == 0)
200 
201                 {
202 
203                   ans1 = ans1|((long long)1<<i);
204 
205                   cnt++;
206 
207                 }
208 
209                 else if(cnt >= 1)
210 
211                 {
212 
213                     ans2 = ans2|((long long)1<<i);
214 
215                     cnt++;
216 
217                 }
218 
219             }
220 
221             else
222 
223             {
224 
225                 if(cnt>1)
226 
227                 {
228 
229                     ans1 = ans1|((long long)1<<i);
230 
231                     ans2 = ans2|((long long)1<<i);
232 
233                 }
234 
235             }
236 
237             //cout<<ans2<<' '<<ans1<<endl;
238 
239         }
240 
241         //cout<<ans2<<' '<<ans1<<endl;
242 
243         if(ans2<l)
244 
245         {
246 
247             if(l%2==0)
248 
249                 printf("1\n2\n%I64d %I64d\n",l,l+1);
250 
251             else printf("1\n2\n%I64d %I64d\n",l+1,l+2);
252 
253         }
254 
255         else printf("0\n3\n%I64d %I64d %I64d\n",ans1,ans2,r);
256 
257     }
258 
259     return 0;
260 
261 }

 

转载于:https://www.cnblogs.com/haohaooo/p/4035316.html