题目链接:
Series 1
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 423 Accepted Submission(s): 146
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.
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)
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
2 3 1 2 3 4 1 5 7 2
Sample Output
0 -5
Source
题目分析:
公式:C(n-1,0)*a[n]-C(n-1,1)*a[n-1]+C(n-1,2)*a[n-2]+...+C(n-1,n-1)*a[n]。
用C(n,k)=C(n,k-1)*(n-k+1)/k就可以高速得到一行的二项式系数。
公式:C(n-1,0)*a[n]-C(n-1,1)*a[n-1]+C(n-1,2)*a[n-2]+...+C(n-1,n-1)*a[n]。
用C(n,k)=C(n,k-1)*(n-k+1)/k就可以高速得到一行的二项式系数。
第一道Java题!
代码例如以下:
import java.util.Scanner; import java.math.BigInteger; public class Main { public static void main(String[] args) { Scanner cin = new Scanner(System.in); BigInteger[] a; a = new BigInteger[3017]; int t, n; t = cin.nextInt(); while(t--){ n = cin.nextInt(); for (int i = 0; i < n; i++) a[i] = cin.nextBigInteger(); BigInteger ans = BigInteger.ZERO; BigInteger c = BigInteger.ONE; for (int i = 0; i < n; i++) { BigInteger tmp = c.multiply(a[n-i-1]); if (i%2 == 0) ans = ans.add(tmp); else ans = ans.subtract(tmp); tmp = c.multiply(BigInteger.valueOf(n-i-1)); c = tmp.divide(BigInteger.valueOf(i+1)); } System.out.println(ans); } } }