51nod1832(二叉树/高精度模板+dfs)

题目链接: http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1832

题意: 中文题诶~

思路: 若二叉树中有 k 个节点只有一个子树, 则答案为 1 << k.

详情参见:http://blog.csdn.net/gyhguoge01234/article/details/77836484

代码:

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

 const int MAX = 1e2;
 const int M = 1e9;//1e9为1节
 const int MAXN = ;

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

 // 先序遍历 X L … R …
 // 后序遍历 … L … R X

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

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

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