Question

我试图打印出Fibonacci序列，但输出与预期输出不匹配。我使用了printf函数并引用了存储数据的数组，因此我不确定我的错误在哪里。我也得到了奇怪的输出。

如果m = 0，应该打印序列的循环被设计为打印第i个斐波纳契数，应该打印出每个斐波纳契数的8位数。对于任何其他m，它为每个m在数组中存储n个更多空格，并且打印功能被设计为相应地打印。

当m = 0时，只打印前9个Fibbonacci数字，当m = 1时，只有前7个数字显示正确的位数。

我不确定我的错误在哪里。

代码：

#include <stdio.h>
#include <stdlib.h>

int main() {

    int a, b, n;

    //create a list for the Fibonacci sequence
    printf("Enter n.\n");
    scanf("%d", &n);

   //store a long for each 8 places

    int m;
    printf("Enter level of precision.");
    scanf("%d",&m);
    long fibonacci[(m+1)*n];
    fibonacci[0] = 0;
    fibonacci[1] = 1;
    for (int u = 2; u<((m+1)*n-1);u++) {
        fibonacci[u] = 0;
    }
    printf("F_0 =%d\n",fibonacci[0]);
    printf("F_1 =%d\n",fibonacci[1]);
    int j,l;
    for (int i = 2; i<n; i++) {
        a = i-1;
        b = i-2;
        fibonacci[i] = fibonacci[a] + fibonacci[b];
        fibonacci[n+i] = (fibonacci[i]-(fibonacci[i]%100000000))/100000000;
        fibonacci[i] = fibonacci[i]%100000000;

        for (j = 0;j<m;j++) {
            fibonacci[(j+1)*n+i] = fibonacci[(j+1)*n+i] + fibonacci[(j+1)*n+a] + fibonacci[(j+1)*n+b];
            fibonacci[(j+2)*n+i] = (fibonacci[(j+1)*n+i] - (fibonacci[(j+1)*n+i]%100000000)) / 100000000;
            fibonacci[(j+1)*n+i] = fibonacci[(j+1)*n+i]%100000000;
        }
        printf("F_%d=",i);

        for (l = 0;l<(m+1);l++) {
            printf("%08d",fibonacci[(m-l)*n+i]);
        }
        printf("\n");
    }
}

输出：

对于m = 0

Enter n.
20
Enter level of precision.0
F_0 =0
F_1 =1
F_2=00000001
F_3=00000002
F_4=00000003
F_5=00000005
F_6=00000008
F_7=00000013
F_8=00000021
F_9=00000034

表示m = 1：

Enter n.
20
Enter level of precision.1
F_0 =0
F_1 =1
F_2=0000000000000001
F_3=0000000000000002
F_4=0000000000000003
F_5=0000000000000005
F_6=0000000000000008
F_7=0000000000000013
F_8=00000021
F_9=00000034
F_10=00000055
F_11=00000089
F_12=00000144
F_13=00000233
F_14=00000377
F_15=00000610
F_16=00000987
F_17=00001597
F_18=00002584
F_19=00004181

斐波那契序列省略数字c

0 个答案: