c语言，编程实现，求斐波那契数列，1,1,2,3,5,8,......的前20项及前20项和 ，c语言：利用数组求斐波那契数列的前20项
\u7528C\u8bed\u8a00\u7f16\u7a0b \u6590\u6ce2\u90a3\u5951\u6570\u5217\uff1a1\uff0c1\uff0c2\uff0c3\uff0c5\uff0c8\uff0c13\u2026\uff0c\u8f93\u51fa\u524d13\u9879\uff0c\u6bcf\u884c\u8f93\u51fa5\u4e2a\u6570\u3002public class FeiBoMethod {
// \u4f7f\u7528\u9012\u5f52\u65b9\u6cd5
private static int getSum(int num) {
if (num== 1 || num== 2)
return 1;
else
return getSum(num- 1) + getFibo(num- 2);
}
public static void main(String[] args) {
System.out.println("\u6590\u6ce2\u90a3\u5951\u6570\u5217\u7684\u524d13\u9879\u4e3a\uff1a");
for (int i = 1; i <= 13; i++) {
System.out.print(getSum(i) + "\t");
if (i % 5 == 0)
System.out.println();
}
}
}
\u6269\u5c55\u8d44\u6599\uff1a\u4f7f\u7528\u6570\u7ec4\u7684\u65b9\u5f0f\u5b9e\u73b0
publicclassFeiBoMethod\uff5b
\uff0f\uff0f\u5b9a\u4e49\u6570\u7ec4\u65b9\u6cd5
publicstaticvoidmain\uff08String\uff3b\uff3dargs\uff09\uff5b
intarr\uff3b\uff3d\uff1dnewint\uff3b13\uff3d\uff1b
arr\uff3b0\uff3d\uff1darr\uff3b1\uff3d\uff1d1\uff1b
for\uff08inti\uff1d2\uff1bi\uff1carr\uff0elength\uff1bi\uff0b\uff0b\uff09\uff5b
arr\uff3bi\uff3d\uff1darr\uff3bi\uff0d1\uff3d\uff0barr\uff3bi\uff0d2\uff3d\uff1b
\uff5d
System\uff0eout\uff0eprintln\uff08\uff02\u6590\u6ce2\u90a3\u5951\u6570\u5217\u7684\u524d13\u9879\u5982\u4e0b\u6240\u793a\uff1a\uff02\uff09\uff1b
for\uff08inti\uff1d0\uff1bi\uff1carr\uff0elength\uff1bi\uff0b\uff0b\uff09\uff5b
if\uff08i\uff055\uff1d\uff1d0\uff09
System\uff0eout\uff0eprintln\uff08\uff09\uff1b
System\uff0eout\uff0eprint\uff08arr\uff3bi\uff3d\uff0b\uff02\uff3ct\uff02\uff09\uff1b
\uff5d
\uff5d
\uff5d
\u4ee3\u7801\u5982\u4e0b\uff1a
int a[20]={1\uff0c1}\uff1b
printf\uff08\u201c%d\t%d\t\u201d\uff0ca[0]\uff0ca[1]\uff09\uff1b
for\uff08int i=0\uff1bi<20\uff1bi++\uff09
{
printf\uff08\u201c%d\t\u201d\uff0ca[i]=a[i-1]+a[i-2]\uff09\uff1b
}
\u6269\u5c55\u8d44\u6599\uff1a
\u5982\u4f55\u638c\u63e1C\u8bed\u8a00\u57fa\u7840\u77e5\u8bc6\u53ca\u6280\u5de7\uff1a
\u9996\u5148\u51c6\u5907\u4e00\u53f0PC\uff0c\u5b66\u7f16\u7a0b\u4e00\u822c\u5bf9PC\u6ca1\u6709\u592a\u5927\u7684\u8981\u6c42\uff0c\u4e00\u822c\u4e3b\u6d41\u7684\u914d\u7f6e\u5c31\u597d\u3002
\u6709\u4e86\u7535\u8111\u4e4b\u540e\uff0c\u5c31\u5f97\u8003\u8651\u5b89\u88c5\u4ec0\u4e48\u64cd\u4f5c\u7cfb\u7edf\u4e86\uff0c\u4e3b\u6d41\u4f7f\u7528\u7684\u64cd\u4f5c\u7cfb\u7edf\u662fwindows\uff0c\u4e0d\u8fc7\u5728\u8fd9\u91cc\u4e0d\u5efa\u8bae\u5b66\u4e60C\u7f16\u7a0b\u4f7f\u7528windows\uff0c\u5efa\u8bae\u4f7f\u7528Linux\u3002\u6240\u4ee5\u5b66\u4e60\u8005\u6700\u597d\u5b89\u88c5\u53cc\u7cfb\u7edf\uff0c\u6216\u8005\u5f7b\u5e95\u4e60\u60efLinux\u7cfb\u7edf\u3002
\u4e0d\u7ba1\u5b66\u4e60\u54ea\u95e8\u7f16\u7a0b\u8bed\u8a00\uff0c\u90fd\u5efa\u8bae\u5b66\u4e60\u8005\u5fc5\u987b\u6709\u4e00\u672c\u5173\u4e8e\u6b64\u8bed\u8a00\u7684\u5168\u9762\u77e5\u8bc6\u7684\u4e66\u7c4d\uff0c\u5927\u4e00\u822c\u91c7\u7528\u7684\u662f\u8c2d\u6d69\u5f3a\u8001\u5e08\u7684\u300aC\u8bed\u8a00\u7a0b\u5e8f\u8bbe\u8ba1\u300b\uff0c\u8fd8\u6709 \u300aC\u8bed\u8a00C++\u5b66\u4e60\u6307\u5357\u300b\u4ece\u5165\u95e8\u5230\u7cbe\u901a\uff08\u8bed\u6cd5\u7bc7\uff09\u89c6\u9891\u8bfe\u7a0b\u5728\u7ebf\u5b66\u4e60\u7b49\uff0c\u5efa\u8bae\u5b66\u4e60\u8005\u53ef\u4ee5\u53bb\u4ed4\u7ec6\u7814\u8bfb
C语言源程序如下：
#include<stdio.h>
int main()
{
int array[100]={1,1};//斐波那契数列前两个元素均为0
int i=0;//循环变量
int n=20;//数列需要求的个数
int sum = 0;//和变量
for(i=2;i<n+1;i++)//按递推原理依次求出后续元素
{
array[i]=array[i-1]+array[i-2];//数列原理
}
printf("arr[1]-arr[%d] = ", n);//提示输出数列元素
for (i = 0; i < n; i++)//遍历数列
{
printf("%d ",array[i]);//输出arr[1]-arr[n]元素内容
sum += array[i];//顺便进行求和
}
printf("
%d ", sum);//输出求和结果
return 0;
}
程序运行结果如下：
扩展资料：
利用递归原理进行求斐波那契数列的求解和求前n项和：
#include<stdio.h>
int fibonacci(int n) //递归函数
{
if (n == 0 || n == 1)
return 1;
if (n > 1)
return fibonacci(n - 1) + fibonacci(n - 2);
}
int main()
{
int i = 0;
for (i = 0; i < 30; i++)
{
printf("fibonacci(%d) = %d
", i, fibonacci(i));
}
return 0;
}
#include<stdio.h>
int main()
{int a[20],n,i,s=2;
a[0]=a[1]=1;
printf("1 1 ");
for(i=2;i<20;i++)
{a[i]=a[i-1]+a[i-2];
s+=a[i];
printf("%d ",a[i]);
}
printf("
sum=%d
",s);
return 0;
}
#include<stdio.h>
int fib(int n);
int main(void)
{
int sum=0;
for(int i=1;i<=20;i++)
{
if(i%11==0)
printf("
");
printf("%5d",fib(i));
sum+=fib(i);
}
printf("
前20项之和为：%d
",sum);
}
int fib(int n)
{
if(n<3)
return 1;
else
return fib(n-1)+fib(n-2);
}
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绛旓細public class FeiBoMethod { // 浣跨敤閫掑綊鏂规硶 private static int getSum(int num) { if (num== 1 || num== 2)return 1;else return getSum(num- 1) + getFibo(num- 2);} public static void main(String[] args) { System.out.println("鏂愭尝閭ｅ鏁板垪鐨勫墠13椤逛负锛");for (int i ...
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绛旓細include <stdio.h> int main(){ int f1 = 1;int f2 = 1;int f3,i;printf("%d\t%d\t",f1,f2);for(i = 1; i <= 38; i++){ f3 = f1 + f2;printf("%d\t",f3);f1= f2;f2= f3;} printf("\n");return 0;}
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