This question requires the preparation of program that calculates the sequence 2/1+3/2+5/3+8/5+...of front Nand Paragraph. Note that the sequence starts from the first 2item, the numerator of each item is the sum of the previous item's numerator and the denominator, and the denominator is the previous item's numerator.
Input format:
Enter a positive integer in one line N.
Output format:
Output the partial sum value in one line, accurate to two decimal places. The title guarantees that the calculation result does not exceed the double precision range.
Input sample:
20
Sample output:
32.66
Code:
# include <stdio.h>
# include <stdlib.h>
int main() {
int N;
scanf("%d",&N);
double value = 0.0,a = 1.0,b = 2.0,temp;
// a和b分别代表分母和分子
int i = 1;
while (i <= N) {
value += (b / a);
// 第(i+1)分母 == 第i分子，第(i+1)分子 = 第i分母 + 第i分子
temp = b;
b += a;
a = temp;
i += 1;
}
printf("%.2lf",value);
return 0;
}
Submit screenshot:
Problem-solving ideas:
The mathematical idea of this question is Fibonacci 纳妾(O(∩_∩)O). It was originally used to study the problem of rabbit reproduction. Now it is a PATpractice problem to change it ! The following divides the fraction into numerator and denominator to understand:
- molecular:
2 3 5 8 - Denominator:
1 2 3 5
We can find out第(i+1)分母 == 第i分子，第(i+1)分子 = 第i分母 + 第i分子, so we can easily write the code!