51nod 1832 先序遍历与后序遍历（dfs+高精度）

Posted 2020-10-17 谦谦君子，陌上其华

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

题意：

 

思路：

官方题解如下：

可以看一下这篇文章：https://wenku.baidu.com/view/a2a45aa0284ac850ad024261.html

  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<queue>
  5 using namespace std;
  6 const int maxn = 10000+5;
  7 const int inf = 0x3f3f3f3f;
  8 
  9 int n;
 10 int a[maxn],b[maxn];
 11 
 12 struct BigInt
 13 {
 14     const static int mod = 10000;
 15     const static int DLEN = 4;
 16     int a[600],len;
 17     BigInt()
 18     {
 19         memset(a,0,sizeof(a));
 20         len = 1;
 21     }
 22     BigInt(int v)
 23     {
 24         memset(a,0,sizeof(a));
 25         len = 0;
 26         do
 27         {
 28             a[len++] = v%mod;
 29             v /= mod;
 30         }while(v);
 31     }
 32     BigInt(const char s[])
 33     {
 34         memset(a,0,sizeof(a));
 35         int L = strlen(s);
 36         len = L/DLEN;
 37         if(L%DLEN)len++;
 38         int index = 0;
 39         for(int i = L-1;i >= 0;i -= DLEN)
 40         {
 41             int t = 0;
 42             int k = i - DLEN + 1;
 43             if(k < 0)k = 0;
 44             for(int j = k;j <= i;j++)
 45                 t = t*10 + s[j] - \'0\';
 46             a[index++] = t;
 47         }
 48     }
 49     BigInt operator +(const BigInt &b)const
 50     {
 51         BigInt res;
 52         res.len = max(len,b.len);
 53         for(int i = 0;i <= res.len;i++)
 54             res.a[i] = 0;
 55         for(int i = 0;i < res.len;i++)
 56         {
 57             res.a[i] += ((i < len)?a[i]:0)+((i < b.len)?b.a[i]:0);
 58             res.a[i+1] += res.a[i]/mod;
 59             res.a[i] %= mod;
 60         }
 61         if(res.a[res.len] > 0)res.len++;
 62         return res;
 63     }
 64     BigInt operator *(const BigInt &b)const
 65     {
 66         BigInt res;
 67         for(int i = 0; i < len;i++)
 68         {
 69             int up = 0;
 70             for(int j = 0;j < b.len;j++)
 71             {
 72                 int temp = a[i]*b.a[j] + res.a[i+j] + up;
 73                 res.a[i+j] = temp%mod;
 74                 up = temp/mod;
 75             }
 76             if(up != 0)
 77                 res.a[i + b.len] = up;
 78         }
 79         res.len = len + b.len;
 80         while(res.a[res.len - 1] == 0 &&res.len > 1)res.len--;
 81         return res;
 82     }
 83     void output()
 84     {
 85         printf("%d",a[len-1]);
 86         for(int i = len-2;i >=0 ;i--)
 87             printf("%04d",a[i]);
 88         printf("\\n");
 89     }
 90 };
 91 BigInt ans;
 92 
 93 void dfs(int al, int ar, int bl, int br)
 94 {
 95     if(ar-al<=0)  return;
 96     al++;
 97     br--;
 98     int cnt = 0;
 99     int index = bl;
100     while(a[al]!=b[index])  index++;
101     int newar = al+index-bl+1;
102     int newbl = index+1;
103     cnt++;
104     dfs(al,newar-1,bl,index);
105     if(ar-al!=index-bl)
106     {
107         cnt++;
108         dfs(newar,ar,newbl,br);
109     }
110     if(cnt==1)  ans=ans*2;
111 }
112 
113 int main()
114 {
115     //freopen("in.txt","r",stdin);
116     ans = 1;
117     scanf("%d",&n);
118     for(int i=0;i<n;i++)  scanf("%d",&a[i]);
119     for(int i=0;i<n;i++)  scanf("%d",&b[i]);
120     dfs(0,n-1,0,n-1);
121     ans.output();
122     return 0;
123 }