If use Dinis -> remember that Oriented Graph

thanks very much for your advice

Than you: for test 2 3 answer 0

"/*
const int maxn = 100100;

int a[maxn], b[maxn], c[maxn];
bool order[maxn];
int cnt;
int p[maxn], rank[maxn];


int find_set(int x) {
    if(p[x] == x) return x;
    return p[x] = find_set(p[x]);
}

int unite(int x, int y) {
    x = find_set(x);
    y = find_set(y);
    if(x == y)
        return 0;
    if(rank[x] > rank[y]) {
        p[y] = x;
    } else {
        p[x] = y;
        if(rank[x] == rank[y])
            rank[y] ++;
    }
    return 1;
}

int main() {

    int n, m, q;
    scanf("%d%d", &n, &m);
    for (int i = 0; i < m; i ++) {
        scanf("%d%d", a + i, b + i);
    }
    scanf("%d", &q);
    for (int i = 0; i < q; i ++) {
        scanf("%d", c + i);
        order[--c[i]] = true;
    }
    for (int i = 0; i < n; i ++)  p[i] = i;
    cnt = n;
    for (int i = 0; i < n; i ++) {
        if(order[i] == false) {
            cnt -= unite(a[i], b[i]);
        }
    }

    c[q] = cnt;
    for (int i = q - 1; i > 0; i --) {
        cnt -= unite(a[c[i]], b[c[i]]);
        c[i] = cnt;
    }
    for (int i = 1; i < q; i ++)
        printf("%d ", c[i]);
    printf("%d", c[q]);
}*/"

It's not true. gvsmirnov meant that you can get on input team of 1 person

Edited by author 19.05.2020 20:44

Edited by author 19.05.2020 20:44

You are right!

Should be 11 though, I don't know why authors decided on 1. The solution requires a product of digits. 1 by itself isn't a product at all.

Edited by author 19.05.2020 18:51

I have WA40 !

Don't forget rotate clockwise and anticlockwise at one step!