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HDU 4927 Series 1 ( 组合+高精度)

2024-07-17 02:17:55   219人阅读

Series 1

 

大意：

题意不好翻译，英文看懂也不是很麻烦，就不翻译了。

Problem Description
Let A be an integral series {A1, A2, . . . , An}.

The zero-order series of A is A itself.

The first-order series of A is {B1, B2, . . . , Bn-1},where Bi = Ai+1 - Ai.

The ith-order series of A is the first-order series of its (i - 1)th-order series (2<=i<=n - 1).

Obviously, the (n - 1)th-order series of A is a single integer. Given A, figure out that integer.
 

 

Input
The input consists of several test cases. The first line of input gives the number of test cases T (T<=10).

For each test case:
The first line contains a single integer n(1<=n<=3000), which denotes the length of series A.
The second line consists of n integers, describing A1, A2, . . . , An. (0<=Ai<=105)
 

 

Output
For each test case, output the required integer in a line.
 

 

Sample Input
231 2 341 5 7 2
 

 

Sample Output
0-5
 
 
思路：
比赛中楠姐很快就推出来公式了，想把杨辉三角预处理出来，然后发现BigInteger大小爆内存了。。。。很无语
然后又想在暴力的基础上去优化，然后一直T到死。。。 比赛结束也没搞出来。
赛后才知道，杨辉三角是可以直接用组合公式推出来的。。。
杨辉三角的第n行的第m个数为组合数c[n-1][m-1]。
应用c[n][m] = c[n][m-1]*(n-m+1)/m，就可以快速递推出第n行的数，这样既避免了打表会出现的爆内存，也省去了暴力好多的时间。。。。。
还是太年轻  哎。。。
 
 1 import java.io.*; 2 import java.math.*; 3 import java.util.*; 4 public class Main {             5      6     static BigInteger coe[][] = new BigInteger [3010][3010]; 7     public static void main(String[] args) throws IOException{ 8         Scanner cin = new Scanner(System.in); 9         BigInteger []a = new BigInteger[3010];  10         BigInteger []c = new BigInteger[3010]; 11         int T;12         T = cin.nextInt();13         while(T-- > 0){14             int n;15             n = cin.nextInt();16             for(int i = 1; i <= n; ++i){17                 a[i] = cin.nextBigInteger();18             }19             BigInteger ans = BigInteger.ZERO;20             c[0] = BigInteger.ONE;21             ans = ans.add(c[0].multiply(a[n]));22             int t = -1;23             for(int i = 1; i < n; ++i){24                 BigInteger t1 = BigInteger.valueOf(n).subtract(BigInteger.valueOf(i));  25                 BigInteger t2 = BigInteger.valueOf(i);26                 c[i] = c[i-1].multiply(t1).divide(t2); 27                 ans = ans.add(c[i].multiply(a[n-i]).multiply(BigInteger.valueOf(t)));28                 t *= -1; 29             }30             System.out.println(ans);31         }32         33     }34 }
HDU 4927

 

 


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