In Arcady‘s garden there grows a peculiar apple-tree that fruits one time per year. Its peculiarity can be explained in following way: there are n inflorescences, numbered from 1 to n. Inflorescence number 1 is situated near base of tree and any other inflorescence with number i (i?>?1) is situated at the top of branch, which bottom is p

Once tree starts fruiting, there appears exactly one apple in each inflorescence. The same moment as apples appear, they start to roll down along branches to the very base of tree. Each second all apples, except ones in first inflorescence simultaneously roll down one branch closer to tree base, e.g. apple in a

Help Arcady with counting number of apples he will be able to collect from first inflorescence during one harvest.

First line of input contains single integer number n (2?≤?n?≤?100?000) — number of inflorescences.

Second line of input contains sequence of n?-?1 integer numbers p₂,?p₃,?...,?p_n (1?≤?p_i?<?i), where p_i is number of inflorescence into which the apple from i-th inflorescence rolls down.

Single line of output should contain one integer number: amount of apples that Arcady will be able to collect from first inflorescence during one harvest.

3
1 1

1

5
1 2 2 2

3

18
1 1 1 4 4 3 2 2 2 10 8 9 9 9 10 10 4

4

In first example Arcady will be able to collect only one apple, initially situated in 1st inflorescence. In next second apples from 2nd and 3rd inflorescences will roll down and annihilate, and Arcady won‘t be able to collect them.

In the second example Arcady will be able to collect 3 apples. First one is one initially situated in first inflorescence. In a second apple from 2nd inflorescence will roll down to 1st (Arcady will collect it) and apples from 3rd, 4th, 5th inflorescences will roll down to 2nd. Two of them will annihilate and one not annihilated will roll down from 2-nd inflorescence to 1st one in the next second and Arcady will collect it.

題意 ： 給你一顆樹，樹上共有 n 個結點，在每個時刻樹上的蘋果會向下脫落，但是落在同一節點的蘋果數如果是偶數的話，則會全部爛掉，如果是奇數的話則會剩下一個，問所有蘋果落至根節點後會剩下多少個。

思路分析 ： 計算出每層的蘋果數量，對2取余，累加即可。

代碼示例 ：

const int maxn = 1e5+5;
const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;

int a[maxn], cnt[maxn];

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int n, x;
    
    cin >> n;
    a[1] = 1; cnt[1] = 1;
    for(int i = 2; i <= n; i++){
        scanf("%d", &x);
        a[i] = a[x] + 1;
        cnt[a[i]]++;
    }
    int ans = 0;
    for(int i = 1; i <= n; i++){
        if (cnt[i] % 2) ans++;
    }
    printf("%d\n", ans);
    return 0;
}

思維題