#include <stdlib.h>
#include <stdio.h>
int main (){
int n, cont, fib, na = 0, nb = 1, sum_even = 0;
printf ("Insert a number and I'll tell you the respective Fibonacci: ");
scanf ("%d", &n);
for (cont = 1; cont < n; cont++) {
na += nb;
nb = na - nb;
fib = na + nb;
if (fib % 2 == 0) {
sum_even += fib;
}
}
printf ("%d\n", sum_even);
return 0;
}
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我试图做项目欧拉问题2,然后我想出了这个代码.问题是:我找不到斐波那契序列中对数的总和,因为数字超过400或接近数字,因为内存溢出.因此,我无法解决这个练习,因为它要求在斐波那契序列中找到4000000以下的对数的总和.谁能帮我?
编辑:我试图使用浮点数类型来增加答案的容量,它似乎工作到一千左右,但如果我尝试更大的数字,我在bash中得到一个-nan错误,像处理15秒(我不真的知道这意味着什么.
#include <stdlib.h>
#include <stdio.h>
int main () {
int n, cont, div;
float sum_even = 0, na = 0, nb = 1, fib;
printf ("Insert a number and I'll tell you the respective Fibonacci: ");
scanf ("%d", &n);
for (cont = 1; cont <= n; cont++) {
na += nb;
nb = na - nb;
fib = na + nb;
div = fib / 2;
if (div % 2 == 0) {
sum_even += fib;
}
}
printf ("%f\n", sum_even);
return 0;
}
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