HDU1134 Game of Connections 高精度+DP

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Problem Address：http://acm.hdu.edu.cn/showproblem.php?pid=1134

【思路】

dp状态转移：dp[i] = sum(dp[j]*dp[i-1-j]) [0<=j<i]

加上高精度乘法和加法。

【代码】

代码又颓了= =

看来得准备一个高精度的模板啊！

#include <iostream>using namespace std;const int maxn = 100;const int size = 100;  struct num  {      int n[size];      int len;      num()      {          memset(n, 0, sizeof(n));          len = 1;      }  }d[maxn+5];void multiply(num &t, num s){      int i,j;      num a;      for (i=0; i<t.len; i++)      {          for (j=0; j<s.len; j++)          {              a.n[i+j] += t.n[i] * s.n[j];          }      }      for (i=0; i<size-1; i++)      {          if (a.n[i]>=10)          {              a.n[i+1] += a.n[i]/10;              a.n[i] = a.n[i]%10;          }      }      for (i=size-1; i>=0; i--)      {          if (a.n[i]!=0) break;      }     a.len = i+1;    if (a.len==0)          a.len = 1;    for (i=0; i<size; i++)      {          t.n[i] = a.n[i];      }      t.len = a.len;  }  void add(num &t, num s){int i;num a;for (i=0; i<size-1; i++){a.n[i] += t.n[i] + s.n[i];}    for (i=0; i<size-1; i++)      {          if (a.n[i]>=10)          {              a.n[i+1] += a.n[i]/10;              a.n[i] = a.n[i]%10;          }      }      for (i=size-1; i>=0; i--)      {          if (a.n[i]!=0) break;      }     a.len = i+1;    if (a.len==0)          a.len = 1;    for (i=0; i<size; i++)      {          t.n[i] = a.n[i];      }      t.len = a.len;  }void output(num a) { int i;for (i=a.len-1; i>=0; i--) printf("%d", a.n[i]);printf("\n");}  int main(){int i, j;d[0].n[0] = 1;d[0].len = 1;d[1].n[0] = 1;d[1].len = 1;for (i=2; i<=maxn; i++){num temp;for (j=0; j<i; j++){temp = d[j];multiply(temp, d[i-1-j]);add(d[i], temp);}}int n;while(scanf("%d", &n)!=EOF){if (n==-1) break;output(d[n]);}return 0;}