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(" "); 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 }