标签：different logs which not lin using 计算 解法 --
InputEach line of the input file will be a single positive number n, except the last line, which is a number -1. You may assume that 1 <= n <= 100.
OutputFor each n, print in a single line the number of ways to connect the 2n numbers into pairs.
Sample Input
2
3
-1
Sample Output
2 5
解法:
卡特兰数的计算公式:a[n] = a[n-1]*(4*n-1)/n+1;
主要的问题在于大数的处理上
1 #include<iostream>
2 #include<cstdio>
3 #include<algorithm>
4 #include<cstring>
5
6 using namespace std;
7
8 const int N = 10000;
9 int a[101][101];
10
11 void cheng( int temp ,int i)
12 {
13 int temp1 = 0;
14 for(int j = 100;j >= 0;j--)
15 {
16 temp1 = a[i-1][j] * temp + temp1;
17 a[i][j] = temp1 % N;
18 temp1 = temp1 / N;
19 }
20
21 }
22
23 void chu (int temp,int i)
24 {
25 int temp1 = 0;
26 for(int j = 0;j <= 100;j++)
27 {
28 temp1 = temp1*N + a[i][j];
29 a[i][j] = temp1 / temp;
30 temp1 = temp1 % temp;
31 }
32
33 }
34
35
36 void DP()
37 {
38 memset(a,0,sizeof(a));
39 a[1][100] = 1;
40 for(int i = 2;i <= 100;i++)
41 {
42 int temp;
43 temp = 4*i-2;
44 cheng(temp,i);
45 temp = i + 1;
46 chu(temp,i);
47 }
48
49 }
50
51 int main ()
52 {
53 DP();
54 int t;
55 while(cin>>t)
56 {
57 if(t == -1)
58 break;
59 int temp;
60 for(int i = 0;i <=100;i++)
61 if( a[t][i] )
62 {
63 temp = i;
64 break;
65 }
66 printf("%d",a[t][temp]);
67 for(int i = temp+1;i <= 100;i++){
68 printf("%04d",a[t][i]);
69 }
70 cout<<endl;
71 }
72 }
标签：different logs which not lin using 计算 解法 --
原文地址：http://www.cnblogs.com/a2985812043/p/7220546.html
