标签：break output cst while ++ printf mat out 3.1
Make Triangle
Chayanika loves Mathematics. She is learning a new chapter geometry. While reading the chapter a question came in her mind. Given a convex polygon of n sides. In how many ways she can break it into triangles, by cutting it with (n-3) non-adjacent diagonals and the diagonals do not intersect.
First line of the input will be an integer t (1<=t<=100000) which is the no of test cases. Each test case contains a single integer n (3<=n<=1000) which is the size of the polygon.
For each test case output the no of ways %100007.
Input: 2
3
5
Output: 1
5
很迷……答案显然就是卡特兰数
然后在计算的时候出现了一些小小的偏差
c[i]=c[i-1]*(4i-2)/(i+1)的递推式是不行的，因为特么取模的1e5+7不是质数，i+1的逆元怎么搞啊……
然后用n^2的c[k]*c[i-k]的递推了
1 #include<cstdio> 2 #include<iostream> 3 #include<cstring> 4 #include<cstdlib> 5 #include<algorithm> 6 #include<cmath> 7 #include<queue> 8 #include<deque> 9 #include<set> 10 #include<map> 11 #include<ctime> 12 #define LL long long 13 #define inf 0x7ffffff 14 #define pa pair<int,int> 15 #define mkp(a,b) make_pair(a,b) 16 #define pi 3.1415926535897932384626433832795028841971 17 #define mod 100007 18 using namespace std; 19 inline LL read() 20 { 21 LL x=0,f=1;char ch=getchar(); 22 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} 23 while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} 24 return x*f; 25 } 26 LL catlan[1010]; 27 int main() 28 { 29 catlan[1]=catlan[0]=1; 30 for (int i=2;i<=1000;i++) 31 for (int j=0;j<i;j++) 32 catlan[i]=(catlan[i]+catlan[j]*catlan[i-j])%mod; 33 int T=read(); 34 while (T--)printf("%lld\n",catlan[read()-1]); 35 }
标签：break output cst while ++ printf mat out 3.1
原文地址：http://www.cnblogs.com/zhber/p/7152927.html
