我正在尝试对正递减浮点数的排序数组求和。我发现对它们求和的最佳方法是开始从最低到最高将数字相加。我编写此代码是为了提供一个示例,但是,从最高数字开始的总和更精确。为什么? (当然,1/k^2之和应该是f=1.644934066848226)。
#include <stdio.h>
#include <math.h>
int main() {
double sum = 0;
int n;
int e = 0;
double r = 0;
double f = 1.644934066848226;
double x, y, c, b;
double sum2 = 0;
printf("introduce n\n");
scanf("%d", &n);
double terms[n];
y = 1;
while (e < n) {
x = 1 / ((y) * (y));
terms[e] = x;
sum = sum + x;
y++;
e++;
}
y = y - 1;
e = e - 1;
while (e != -1) {
b = 1 / ((y) * (y));
sum2 = sum2 + b;
e--;
y--;
}
printf("sum from biggest to smallest is %.16f\n", sum);
printf("and its error %.16f\n", f - sum);
printf("sum from smallest to biggest is %.16f\n", sum2);
printf("and its error %.16f\n", f - sum2);
return 0;
}
您的代码创建一个数组double terms[n];
在堆栈上,这对程序崩溃之前可以执行的迭代次数设置了严格限制。
但您甚至没有从该数组中获取任何内容,因此根本没有理由将其放在那里。我修改了你的代码以摆脱terms[]
:
#include <stdio.h>
int main() {
double pi2over6 = 1.644934066848226;
double sum = 0.0, sum2 = 0.0;
double y;
int i, n;
printf("Enter number of iterations:\n");
scanf("%d", &n);
y = 1.0;
for (i = 0; i < n; i++) {
sum += 1.0 / (y * y);
y += 1.0;
}
for (i = 0; i < n; i++) {
y -= 1.0;
sum2 += 1.0 / (y * y);
}
printf("sum from biggest to smallest is %.16f\n", sum);
printf("and its error %.16f\n", pi2over6 - sum);
printf("sum from smallest to biggest is %.16f\n", sum2);
printf("and its error %.16f\n", pi2over6 - sum2);
return 0;
}
当运行十亿次迭代时,最小优先方法要准确得多:
Enter number of iterations:
1000000000
sum from biggest to smallest is 1.6449340578345750
and its error 0.0000000090136509
sum from smallest to biggest is 1.6449340658482263
and its error 0.0000000009999996
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