传送门

给一颗销售供应的树。在树根初货物价格为P，从树根每往子节点走一层，价格会上涨R%。求最高价格和最高价格叶节点个数

思路：最高价格其实=树的深度，进行带深度的遍历即可

#include <iostream>
#include<bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<n;i++)
#define sca(x) scanf("%d",&x)
#define sca2(x,y) scanf("%d%d",&x,&y)
#define scl(x) scanf("%lld",&x)
#define pri(x) printf("%d\n",x)
#define pri2(x,y) printf("%d %d\n",x,y)
#define prs(x) printf("%s\n",(x))
#define prl(x) printf("%lld\n",x)
#define ll long long
#define PII pair<int,int>
#define eps 1e-6
#define inf 1e17
#define INF 0x3f3f3f3f
using namespace std;
const int maxn=1e6+5;
vector<int>child[maxn];
int flag[maxn];
int n,m;
int id,cnt,son;
double ans;
int mx;
double p,r;
void dfs(int root,int depth){
    if(child[root].size()==0){
        double price=p*pow((1+r*0.01),depth);
        if(ans<price){
            ans=price;
            mx=1;
        }
        else if(ans==price){
            mx++;
        }
        return;
    }
    rep(i,0,child[root].size()){
        dfs(child[root][i],depth+1);
    }
}

int main(){
    cin>>n>>p>>r;
    int root=0;
    rep(t,0,n){
        cin>>m;
        if(m!=-1){
            child[m].push_back(t);
        }
        else{
            root=t;
        }
    }
    mx=-1;
    dfs(root,0);
    printf("%.2lf %d\n",ans,mx);
}