[hdu1023]递推

http://acm.hdu.edu.cn/showproblem.php?pid=1023

如果把栈里面的元素个数表示成状态，每一步（共2 * n步）的状态构成的状态序列的种数就是答案，令dp[i][j]表示第i步栈的状态为j的方案数，则有：

dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j + 1]，+1、-1相当于进栈和出栈，需考虑边界条件，详见代码（答案太大，需用大数）：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <cmath>
 #include <algorithm>
 #include <map>
 #include <stack>
 #include <string>
 #include <ctime>
 #include <queue>
 #define mem0(a) memset(a, 0, sizeof(a))
 #define mem(a, b) memset(a, b, sizeof(a))
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define eps 0.0000001
 #define lowbit(x) ((x) & -(x))
 #define memc(a, b) memcpy(a, b, sizeof(b))
 #define x_x(a) ((a) * (a))
 #define LL __int64
 #define DB double
 #define pi 3.14159265359
 #define MD 10000007
 #define INF (int)1e9
 using namespace std;
 struct BigNum{
         #define maxlen 10
         #define memc(a, b) memcpy(a, b, sizeof(b))
         #define mem0(a) memset(a, 0, sizeof(a))
         typedef __int64 Num[maxlen + ];
         Num num;
         char s[maxlen + ];
         BigNum operator+(BigNum num2) {
                 BigNum ans;
                 mem0(ans.num);
                 for(int i = ; i <= maxlen; i++) {
                         ans.num[i] += num[i] + num2.num[i];
                         ans.num[i + ] += ans.num[i] / (int)1e9;
                         ans.num[i] %= (int)1e9;
                 }
                 return ans;
         }
         BigNum operator*(BigNum num2) {
                 BigNum ans;
                 mem0(ans.num);
                 for(int i = ; i <= maxlen; i++) {
                         for(int j = ; j <= maxlen; j++) {
                                 if(i + j -  <= maxlen) {
                                         ans.num[i + j - ] += num[i] * num2.num[j];
                                         ans.num[i + j] += ans.num[i + j - ] / (int)1e9;
                                         ans.num[i + j - ] %= (int)1e9;
                                 }
                         }
                 }
                 return ans;
         }
         void convert() {
                 int len = strlen(s), cnt = ;
                 for(int i = len - ; i >= ; i -= ) {
                         int p = , x = , t = ;
                         while(i - p >=  && p < ) {
                                 x += t * (s[i - p] - '');
                                 p++;
                                 t *= ;
                         }
                         num[++cnt] = x;
                 }
         }
         void inp() {
                 mem0(num);
                 scanf("%s", s);
                 convert();
         }
         void outp() {
                 int p = ;
                 for(int i = maxlen; i >= ; i--) {
                         if(num[i]) {
                                 p = i;
                                 break;
                         }
                 }
                 cout<< num[p];
                 while(--p) {
                         int a[] = {}, x = num[p];
                         for(int i = ; i < ; i++) {
                                 a[i] = x % ;
                                 x /= ;
                         }
                         for(int i = ; i >= ; i--) {
                                 printf("%d", a[i]);
                         }
                 }
         }
         BigNum(char str[]) {
                 strcpy(s, str);
                 mem0(num);
                 convert();
         }
         BigNum(){}
 };
 BigNum f[][];
 int main()
 {
         //freopen("input.txt", "r", stdin);
         //freopen("output.txt", "w", stdout);
         int n;
         while(~scanf("%d", &n)) {
                 mem0(f);
                 f[][] = BigNum("");
                 for(int i = ; i <=  * n; i++) {
                         for(int j = ; j <= n; j++) {
                                 f[i][j] = f[i - ][j + ];
                                 if(j) f[i][j] = f[i][j] + f[i - ][j - ];
                         }
                 }
                 f[ * n][].outp();
                 cout<< endl;
         }
         return ;
 }