51nod1832(二叉樹/高精度模板+dfs)

題目鏈接: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1832

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題意: 中文題誒~

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思路: 若二叉樹中有 k 個節點只有一個子樹, 則答案為 1 << k.

詳情參見:http://blog.csdn.net/gyhguoge01234/article/details/77836484

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代碼:

#include <iostream>
#include <stdio.h>
#include <string.h>
#define ll long long
using namespace std;

const int MAX = 1e2;
const int M = 1e9;//1e9為1節
const int MAXN = 35;

struct BigInt{
    const static int mod = 10000;
    const static int DLEN = 4;
    int a[600], len;
    BigInt(){
        memset(a, 0, sizeof(a));
        len = 1;
    }
    BigInt(int v){
        memset(a, 0, sizeof(a));
        len = 0;
        do{
            a[len++] = v % mod;
            v /= mod;
        }while(v);
    }
    BigInt(const char s[]){
        memset(a, 0, sizeof(a));
        int L = strlen(s);
        len = L / DLEN;
        if(L % DLEN) len++;
        int index = 0;
        for(int i = L - 1; i >= 0; i -= DLEN){
            int t = 0;
            int k = i - DLEN + 1;
            if(k < 0) k = 0;
            for(int j = k; j <= i; j++)
                t = t * 10 + s[j] - '0';
            a[index++] = t;
        }
    }
    BigInt operator +(const BigInt &b)const{
        BigInt res;
        res.len = max(len, b.len);
        for(int i = 0; i <= res.len; i++) res.a[i] = 0;
        for(int i = 0; i < res.len; i++){
            res.a[i] += ((i < len) ? a[i] : 0) + ((i < b.len) ? b.a[i] : 0);
            res.a[i + 1] += res.a[i] / mod;
            res.a[i] %= mod;
        }
        if(res.a[res.len] > 0) res.len++;
        return res;
    }
    BigInt operator *(const BigInt &b)const{
        BigInt res;
        for(int i = 0; i < len; i++){
            int up = 0;
            for(int j = 0; j < b.len; j++){
                int temp = a[i] * b.a[j] + res.a[ i + j] + up;
                res.a[i + j] = temp%mod;
                up = temp / mod;
            }
            if(up != 0)
            res.a[i + b.len] = up;
        }
        res.len = len + b.len;
        while(res.a[res.len - 1] == 0 && res.len > 1) res.len--;
        return res;
    }
    void output(){
        printf("%d", a[len - 1]);
        for(int i = len - 2; i >= 0; i--)
            printf("%04d", a[i]);
        printf("\n");
    }
};

// 先序遍歷 X L … R …
// 后序遍歷 … L … R X

const int N = 1e5 + 10;
int a[N], b[N];
BigInt sol(1);

void dfs(int al, int ar, int bl, int br){
    if(ar - al <= 1) return;
    al++;
    br--;
    int indx = bl, cnt = 0;;
    while(a[al] != b[indx]) indx++;
    int newar = al + (indx - bl + 1);
    int newbr = indx + 1;
    cnt++;
    dfs(al, newar, bl, newbr);
    if(ar - al != indx - bl + 1){
        cnt++;
        dfs(newar, ar, newbr, br);
    }
    if(cnt == 1) sol = sol * 2;
}

int main(void){
    int n;
    scanf("%d", &n);
    for(int i = 0; i < n; i++){
        scanf("%d", &a[i]);
    }
    for(int i = 0; i < n; i++){
        scanf("%d", &b[i]);
    }
    sol = 1;
    dfs(0, n, 0, n);
    sol.output();
    return 0;
}

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