要使用C ++进行XOR最大化，要删除的最小元素数。
问题陈述
给定数字N。任务是找到介于N之间的要删除的最小元素数，以使从其余元素获得的XOR最大。
算法
1. If n is 1 or 2 then there is no need to remove any element. Hence answer is zero 2. Find a number which is power of 2 and greater than or equal to. Let us call this number as nextNumber 2.1. If n == nextNumber or n == (nextNumber – 1) then answer is 1 2.2. If n = (nextNumber -2) then answer is 0 3. If n is an even then answer is 1 otherwise 2
示例
#include <iostream>
using namespace std;
int nextPowerOf2(int n){
if (n && !(n & (n - 1))) {
return n;
}
int cnt = 0;
while (n) {
n = n / 2;
++cnt;
}
return (1 << cnt);
}
int elmentsToBeRemoved(int n){
if (n == 1 || n == 2) {
return 0;
}
int nextNumber = nextPowerOf2(n);
if (n == nextNumber || n == nextNumber -1) {
return 1;
} else if (n == nextNumber - 2) {
return 0;
} else if (n & 1) {
return 2;
} else {
return 1;
}
}
int main(){
int n = 10;
cout << "Numbers to be removed = " <<
elmentsToBeRemoved(n) << endl;
return 0;
}输出结果
当您编译并执行上述程序时。它生成以下输出-
Numbers to be removed = 1s