[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 + 2];
        Num num;
        char s[maxlen + 2];
        BigNum operator+(BigNum num2) {
                BigNum ans;
                mem0(ans.num);
                for(int i = 1; i <= maxlen; i++) {
                        ans.num[i] += num[i] + num2.num[i];
                        ans.num[i + 1] += ans.num[i] / (int)1e9;
                        ans.num[i] %= (int)1e9;
                }
                return ans;
        }
        BigNum operator*(BigNum num2) {
                BigNum ans;
                mem0(ans.num);
                for(int i = 1; i <= maxlen; i++) {
                        for(int j = 1; j <= maxlen; j++) {
                                if(i + j - 1 <= maxlen) {
                                        ans.num[i + j - 1] += num[i] * num2.num[j];
                                        ans.num[i + j] += ans.num[i + j - 1] / (int)1e9;
                                        ans.num[i + j - 1] %= (int)1e9;
                                }
                        }
                }
                return ans;
        }
        void convert() {
                int len = strlen(s), cnt = 0;
                for(int i = len - 1; i >= 0; i -= 9) {
                        int p = 0, x = 0, t = 1;
                        while(i - p >= 0 && p < 9) {
                                x += t * (s[i - p] - '0');
                                p++;
                                t *= 10;
                        }
                        num[++cnt] = x;
                }
        }
        void inp() {
                mem0(num);
                scanf("%s", s);
                convert();
        }
        void outp() {
                int p = 1;
                for(int i = maxlen; i >= 1; i--) {
                        if(num[i]) {
                                p = i;
                                break;
                        }
                }
                cout<< num[p];
                while(--p) {
                        int a[9] = {0}, x = num[p];
                        for(int i = 0; i < 9; i++) {
                                a[i] = x % 10;
                                x /= 10;
                        }
                        for(int i = 8; i >= 0; i--) {
                                printf("%d", a[i]);
                        }
                }
        }
        BigNum(char str[]) {
                strcpy(s, str);
                mem0(num);
                convert();
        }
        BigNum(){}
};
BigNum f[220][120];
int main()
{
        //freopen("input.txt", "r", stdin);
        //freopen("output.txt", "w", stdout);
        int n;
        while(~scanf("%d", &n)) {
                mem0(f);
                f[0][0] = BigNum("1");
                for(int i = 1; i <= 2 * n; i++) {
                        for(int j = 0; j <= n; j++) {
                                f[i][j] = f[i - 1][j + 1];
                                if(j) f[i][j] = f[i][j] + f[i - 1][j - 1];
                        }
                }
                f[2 * n][0].outp();
                cout<< endl;
        }
        return 0;
}