[ NOIP ] Maximum Xor Sum Problem

最新推荐文章于 2025-09-12 18:45:12 发布

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最新推荐文章于 2025-09-12 18:45:12 发布

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文章标签： c/c++

原文链接：http://www.cnblogs.com/Kvar-ispw17/p/6503363.html

本文介绍了一种选择整数序列中某些元素以获得最大异或和的算法。通过从高位到低位扫描，选择合适的元素进行异或操作，最终得到最大的异或结果。

QUESTION

　　Given n integers {a1, a2, ..., an}, n ≤ 100, 000. Select some, s.t. they have the maximum xor sum.

　　For random numbers like 19, 44, 9, 33, they could be made into lines like:

　　　　　　　　　　　　0 1 0 0 1 1

　　　　　　　　　　　　1 0 1 1 0 0

　　　　　　　　　　　　0 0 1 0 0 1

　　　　　　　　　　　　1 0 0 0 0 1

　　 The algorithm is selecting the "1" "0" from high to low order ( left to right ), then from up to down.

　　We need var "int ans" for storing the final answer.

　　Once we find the first "1" in the column (the number we found called "x"), we let the number xor the number below which have got "1" in the same column.

　　Then, we let x xor ans iff ans has "0" at the current column. And we let the number x = 0 0 0 0 0 ......( x = 0 [C++])

　　Loop it in every column till the last one.

　　CODE:

//origin code by cmd2001
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int in[10100], n, ans, x;
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d", in + i);
	x = 0;
	for (int i = 31; i >= 0; i--) {
		x = 0;
		for (int j = 1; j <= n; j++) {
			if ((in[j] >> i) & 1) x = j;
		} if (x) {
			for (int j = 1; j <= n; j++) {
				if (j != x && (in[j] >> i) & 1) in[j] ^= in[x];
			}	
			if (!(ans >> i) & 1) ans ^= in[x];
			in[x] = 0;
		}
	}
	printf("%d\n", ans);
	return 0;
}

转载于:https://www.cnblogs.com/Kvar-ispw17/p/6503363.html