[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）：

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4 #include <cmath>
  5 #include <algorithm>
  6 #include <map>
  7 #include <stack>
  8 #include <string>
  9 #include <ctime>
 10 #include <queue>
 11 #define mem0(a) memset(a, 0, sizeof(a))
 12 #define mem(a, b) memset(a, b, sizeof(a))
 13 #define lson l, m, rt << 1
 14 #define rson m + 1, r, rt << 1 | 1
 15 #define eps 0.0000001
 16 #define lowbit(x) ((x) & -(x))
 17 #define memc(a, b) memcpy(a, b, sizeof(b))
 18 #define x_x(a) ((a) * (a))
 19 #define LL __int64
 20 #define DB double
 21 #define pi 3.14159265359
 22 #define MD 10000007
 23 #define INF (int)1e9
 24 using namespace std;
 25 struct BigNum{
 26         #define maxlen 10
 27         #define memc(a, b) memcpy(a, b, sizeof(b))
 28         #define mem0(a) memset(a, 0, sizeof(a))
 29         typedef __int64 Num[maxlen + 2];
 30         Num num;
 31         char s[maxlen + 2];
 32         BigNum operator+(BigNum num2) {
 33                 BigNum ans;
 34                 mem0(ans.num);
 35                 for(int i = 1; i <= maxlen; i++) {
 36                         ans.num[i] += num[i] + num2.num[i];
 37                         ans.num[i + 1] += ans.num[i] / (int)1e9;
 38                         ans.num[i] %= (int)1e9;
 39                 }
 40                 return ans;
 41         }
 42         BigNum operator*(BigNum num2) {
 43                 BigNum ans;
 44                 mem0(ans.num);
 45                 for(int i = 1; i <= maxlen; i++) {
 46                         for(int j = 1; j <= maxlen; j++) {
 47                                 if(i + j - 1 <= maxlen) {
 48                                         ans.num[i + j - 1] += num[i] * num2.num[j];
 49                                         ans.num[i + j] += ans.num[i + j - 1] / (int)1e9;
 50                                         ans.num[i + j - 1] %= (int)1e9;
 51                                 }
 52                         }
 53                 }
 54                 return ans;
 55         }
 56         void convert() {
 57                 int len = strlen(s), cnt = 0;
 58                 for(int i = len - 1; i >= 0; i -= 9) {
 59                         int p = 0, x = 0, t = 1;
 60                         while(i - p >= 0 && p < 9) {
 61                                 x += t * (s[i - p] - ‘0‘);
 62                                 p++;
 63                                 t *= 10;
 64                         }
 65                         num[++cnt] = x;
 66                 }
 67         }
 68         void inp() {
 69                 mem0(num);
 70                 scanf("%s", s);
 71                 convert();
 72         }
 73         void outp() {
 74                 int p = 1;
 75                 for(int i = maxlen; i >= 1; i--) {
 76                         if(num[i]) {
 77                                 p = i;
 78                                 break;
 79                         }
 80                 }
 81                 cout<< num[p];
 82                 while(--p) {
 83                         int a[9] = {0}, x = num[p];
 84                         for(int i = 0; i < 9; i++) {
 85                                 a[i] = x % 10;
 86                                 x /= 10;
 87                         }
 88                         for(int i = 8; i >= 0; i--) {
 89                                 printf("%d", a[i]);
 90                         }
 91                 }
 92         }
 93         BigNum(char str[]) {
 94                 strcpy(s, str);
 95                 mem0(num);
 96                 convert();
 97         }
 98         BigNum(){}
 99 };
100 BigNum f[220][120];
101 int main()
102 {
103         //freopen("input.txt", "r", stdin);
104         //freopen("output.txt", "w", stdout);
105         int n;
106         while(~scanf("%d", &n)) {
107                 mem0(f);
108                 f[0][0] = BigNum("1");
109                 for(int i = 1; i <= 2 * n; i++) {
110                         for(int j = 0; j <= n; j++) {
111                                 f[i][j] = f[i - 1][j + 1];
112                                 if(j) f[i][j] = f[i][j] + f[i - 1][j - 1];
113                         }
114                 }
115                 f[2 * n][0].outp();
116                 cout<< endl;
117         }
118         return 0;
119 }