A1 = ?
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10248 Accepted Submission(s): 6197
Problem Description
有如下方程:Ai = (Ai-1 + Ai+1)/2 - Ci (i = 1, 2, 3, .... n).
若给出A0, An+1, 和 C1, C2, .....Cn.
请编程计算A1 = ?
Input
输入包括多个测试实例。
对于每个实例,首先是一个正整数n,(n <= 3000); 然后是2个数a0, an+1.接下来的n行每行有一个数ci(i = 1, ....n);输入以文件结束符结束。
Output
对于每个测试实例,用一行输出所求得的a1(保留2位小数).
Sample Input
1
50.00
25.00
10.00
2
50.00
25.00
10.00
20.00
Sample Output
27.50
15.00
思路:
1.找规律
n=1 2A1=A0+A2-2C1
n=2 3A1=2A0+A3-4C1-2C2
n=3 4A1=3A0+A4-6C1-4C2-2C3
n=4 5A1=4A0+A5-8C1-6C2-4C3-2C4
(ps:纯数学,个人A不动,百度了一下,找了一下规律QAQ)
AcCode:
import java.text.DecimalFormat;
import java.util.List;
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
DecimalFormat format = new DecimalFormat("0.00");
while(in.hasNext()) {
int n = in.nextInt();
double a0 = in.nextDouble();
double anAdd1 = in.nextDouble();
double[] nums = new double[n];
for (int i = 0; i < nums.length; i++) {
nums[i] = in.nextDouble();
}
double result = n*a0+anAdd1;
int j = 2*n;
for (int i = 0; i < nums.length; i++) {
result = result - j*nums[i];
j=j-2;
}
System.out.println(format.format(result/(n+1)));
}
}
}