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java - 算术进阶3级

转载 作者:行者123 更新时间:2023-12-01 22:07:35 25 4
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这是算术级数级别 3。输出寻找 2 个下一项。例如,输入为:1, 4, 11, 24。寻找下2项,即45和76。如何解决?

package Function;

import java.util.Scanner;
/**
*
* @author Lenovo
*/

public class ArithmeticProgression {

public static void main(String[] args){
int many;

Scanner keyboard=new Scanner(System.in);
System.out.print("Put many term: ");
many= keyboard.nextInt();
int term[]= new int [many];

int n= 0;

for(int z=0; z<many; z++){
n= n+1;
System.out.format("%d term"+ " is: ", n);
term[z] = keyboard.nextInt();
}

System.out.print("enter the next many terms: ");
int range= keyboard.nextInt();

int term2[] = new int[range+many];

for(int i = 0; i < many; i++){
term2[i] = term[i];
}

int b3= term2[many-1]-term2[many-2];
int b2= term2[many-2]-term2[many-3];
int b1= term2[many-3]-term2[many-4];

int c2= b3-b2;
int c1= b2-b1;

int d= c2-c1;

for(int q=0; q<range; q++){

b3= term2[many-1]-term2[many-2];
b2= term2[many-2]-term2[many-3];
b1= term2[many-3]-term2[many-4];

c2= b3-b2;
c1= b2-b1;

d= c2-c1;

int result= term2[many-1]+b3+c2+d;

System.out.println(result);
many++;

}
}
}

最佳答案

在 q 循环中,您忘记在每一步更新 term2 数组:

    for(int q=0; q<range; q++){

b3= term2[many-1]-term2[many-2];
b2= term2[many-2]-term2[many-3];
b1= term2[many-3]-term2[many-4];

c2= b3-b2;
c1= b2-b1;

d= c2-c1;

int result= term2[many-1]+b3+c2+d;

System.out.println(result);

term2[many] = result; // you forgot this update
many++;
}

这会生成列表 1, 4, 11, 24, 45, 76, 119, 176, 249, 340, 451, ...

关于java - 算术进阶3级,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/58680198/

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