POJ 2084 Game of Connections

/*高精度的Catalan数题目 公式： h ( n ) = h ( n - 1 ) * ( 4 * n - 2 ) / n + 1 */

Source Code

Problem: 2084 User: imutzcyMemory: 296K Time: 0MSLanguage: C++ Result: Accepted

• Source Code

  #include<functional>#include<algorithm>#include<iostream>#include<fstream>#include<sstream>#include<iomanip>#include<numeric>#include<cstring>#include<cassert>#include<cstdio>#include<string>#include<vector>#include<bitset>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<list>#include<set>#include<map>using namespace std;const int MAX=100;const int BASE=10000;int a[101][MAX],i,j,n;void multiply(int *a,int Max,int b)//大数乘法{     int i,array=0;     for(i=Max-1;i>=0;i--){       array+=b*a[i];       a[i]=array%BASE;       array/=BASE;     }} void divide(int *a,int Max,int b)//大数除法{     int i,div=0;     for(i=0;i<=Max-1;i++){       div=div*BASE+a[i];       a[i]=div/b;       div%=b;     }}int main(){    memset(a[1],0,MAX*sizeof(int));    for(i=2,a[1][MAX-1]=1;i<101;i++){      memcpy(a[i],a[i-1],MAX*sizeof(int));    //h[1] = h[i-1]      multiply(a[i],MAX,4*i-2);               //h[i] * = (4 * i - 2);      divide(a[i],MAX,i+1);                   //h[i] / = (i + 1);    }    while(cin>>n,n+1){      for(i=0;i<MAX&&a[n][i]==0;i++);       //去掉数组前为 0 的数字       cout<<a[n][i++];                      //输出第一个非0数       for(;i<MAX;i++)        printf("%04d",a[n][i]);               // %04d  输出后面的数，并每位都保持5位长度       cout<<endl;    }    return 0;}