CF 题目集锦 PART 3 #262 div 2 D_html/css_WEB-ITnose
【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:
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
【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
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;iR) 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) {cout3) { S=(L&1)?L+1:L; cout=L) {cout