Codeforces Round #456 Div.2 B. New Year's Eve

àì夳堔傛蜴生んèń 2022-05-31 10:10 144阅读 0赞

# Problem #

codeforces.com/contest/912/problem/B

# Analysis #

You can choose k or less than k integers from 1~n, calculate maximum xor result  
(1) k = 1, You can only choose one integer, so you choose n and get the maximum xor result n  
(2) k > 1, whatever k is, such as k = 2, or k = 3, ……, or k = n,  
if n has x bits as binary code, the maximum xor result must be 2 ^ x - 1

本题是求从1~n个数中取小于或等于k个数（比如n = 4, k = 3表示从1~4中取1个数或2个数或3个数），对这些数进行异或求和，求最大的异或求和值。  
（1）k = 1时，即从1~n个数中1个数，那么最大的数自然是n。结果为n  
（2）k > 1时，假如n以二进制表示是x位数，则结果必为2 ^ x - 1

### Example 1 ###

n = 4, k = 2  
choose 4 and 3, maximum = 4 xor 3 = 7  
可以取4和3这两个数，最大异或结果 = 4 xor 3 = 7 = 2 ^ 3 - 1

### Example 2 ###

n = 4, k = 3  
Choose two numbers, 4 and 3. Or choose three numbers, 4, 1 and 2.  
maximum = 4 xor 3 = 4 xor 1 xor 2 = 7 = 2 ^ 3 - 1  
可以取4和3这两个数，也可以取4、1、2这三个数。  
最大异或结果 = 4 xor 3 = 4 xor 1 xor 2 = 7 = 2 ^ 3 - 1

### Example 3 ###

n = 4, k = 4  
Choose two numbers, 4 and 3. Or choose three numbers, 4, 1 and 2.  
maximum = 7 = 2 ^ 3 - 1  
You can not choose all for numbers, because 4 xor 1 xor 2 xor 3 xor 4 = 3, which is not the maximum result.  
可以取4和3，也可以取4、1、2。最大异或结果为7。  
但不能取4、1、2、3。因为4 xor 1 xor 2 xor 3 xor 4 = 3，不是最大的结果。

### Example 4 ###

n = 6, k = 6  
Just choose 1 and 6, maximum = 6 xor 1 = 7 = 2 ^ 3 - 1  
只要在1~6中取6和1两个数就行了，6 xor 1 = 7 = 2 ^ 3 - 1

# Code #

### Python2 ###

import sys
    
    n, k = map(int, raw_input().split())
    
    if 1 == k:
        print n
        sys.exit(0)
    
    res = 1
    while res < n:
        res = res * 2 + 1
    
    print res

### C Language ###

#include <stdio.h>
    #include <stdint.h>
    
    int main(void)
    {
        int64_t n, k;
        scanf("%I64d %I64d", &n, &k);
    
        if (1 == k) 
        {
            printf("%I64d\n", n);
    
            return 0;
        }
    
        int bit = 0;
        for (; n >> bit; bit++);
        printf("%d", n);
        printf("%d", bit);
    
        printf("%I64d\n", (1LL << bit) - 1);
    
        return 0;
    }

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