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算術(shù)進(jìn)階3級(jí)

算術(shù)進(jìn)階3級(jí)

烙印99 2024-01-28 17:35:26
這是算術(shù)級(jí)數(shù)級(jí)別 3。輸出尋找 2 個(gè)下一項(xiàng)。例如,輸入為:1, 4, 11, 24。尋找下2項(xiàng),即45和76。如何解決?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++;             }    }}
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慕標(biāo)琳琳

TA貢獻(xiàn)1830條經(jīng)驗(yàn) 獲得超9個(gè)贊

在 q 循環(huán)中,您忘記在每一步更新 term2 數(shù)組:


    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++;     

    }

這會(huì)生成列表 1, 4, 11, 24, 45, 76, 119, 176, 249, 340, 451, ...


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