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Give two arrays $a, b with a length of $n $$

You can replace one of $a $with the sum of all of $xor $at a time.

If $a $array can be converted to $b $array, output the minimum number of operations

Otherwise output $- 1$

Sol

Generally, if you see this kind of $N \leqslant 10^5 $and can't do it, you must find the conclusion first

It's not hard to see that when we replace all the numbers $xor $and then $xor $again, we must get the replaced numbers.

In fact, we can put the number from xor into a new position $N+1$, so that each operation becomes the number of exchanging position $N+1$and any position $x$

And the general problem becomes

First, judge the situation of $- 1 $. Obviously, after sorting two arrays, if there is a position different, there must be no solution

Otherwise, there must be solutions.

I won't be here....

The official explanation is very immortal.

For $I $locations, if $a_i \not = b_i $, then connect $a_i $to $b_i $

Final answer = total number of sides + number of connected blocks - 1

Think about why, for the points in the connection block, if their size is $x $, we can change their corresponding $a $and $b $by $x-1 $operations

For different connecting blocks, we also need one step to convert the number of the $N+1 $positions between the two connecting blocks (except the first one)

For the position of $N+1 $, you need to consider it separately: if it is already in the Unicom block, you don't need to consider it; otherwise, you need to consider it as a separate Unicom block

otherwise

2
1 3
3 1

The number of connecting blocks can be maintained by parallel query sets

#include<bits/stdc++.h>
const int MAXN = 4e5 + 10;
using namespace std;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N;
int a[MAXN], b[MAXN], ta[MAXN], tb[MAXN], sa, sb, tot = 0, date[MAXN], fa[MAXN];
map<int, bool> ti;
int find(int x) {
    return fa[x] == x ? fa[x] : fa[x] = find(fa[x]);
}
int unionn(int x, int y) {
    fa[x] = y;
}
int main() {
    N = read();
    for(int i = 1; i <= N; i++) a[i] = read(), sa ^= a[i]; a[N + 1] = sa;
    for(int i = 1; i <= N; i++) b[i] = read(), sb ^= b[i]; b[N + 1] = sb;
    N++;
    memcpy(ta, a, sizeof(a)); memcpy(tb, b, sizeof(b));
    sort(ta + 1, ta + N + 1); sort(tb + 1, tb + N + 1);
    for(int i = 1; i <= N - 1; i++) if(ta[i] != tb[i]) return puts("-1"), 0;
 
    int ans = 0, num = 0;
    for(int i = 1; i <= N; i++) 
        if(a[i] != b[i] || (i == N)) {
            date[++num] = a[i]; date[++num] = b[i];
            if(i < N)ans++;//The last one is considered separately
        }
    if(ans == 0) return puts("0"), 0;
 
    sort(date + 1, date + num + 1);
    num = unique(date + 1, date + num + 1) - date - 1;
    for(int i = 1; i <= num; i++) fa[i] = i;
    for(int i = 1; i <= N; i++)
        if(a[i] != b[i]) {
            a[i] = lower_bound(date + 1, date + num + 1, a[i]) - date,
            b[i] = lower_bound(date + 1, date + num + 1, b[i]) - date;
            if(!ti[a[i]]) ti[a[i]] = 1;
            if(!ti[b[i]]) ti[b[i]] = 1;
            unionn(find(a[i]), find(b[i]));
        }
        
    for(int i = 1; i <= num; i++)
        if(fa[i] == i) ans++;
    printf("%d", ans - 1);
 
    return 0;
}