2 0 1 8 3 2 3=23加减3

用c语言写s=1/2+2/3+3/5+5/8+....前20项的和_百度知道
用c语言写s=1/2+2/3+3/5+5/8+....前20项的和
我有更好的答案
i = jvoid main(){
int i, j, k, k &=20; k++)
sum += (float)i/(float)j;
j =sum = %f\n&, sum);
for (k = 1
采纳率:34%
include &清空键盘输入缓冲区
scanf(&%d&:////n&);
printf(&///初值 实形数
for(n=0; n&N; n++)
An=An_1+An_2;/累加
An_2=An_1; An_1=An;/////////////&#47, An;//\n请输入累加的前N项数目;///////////叠代
return S;//////////double theSum(int N){// &#47, n, theSum(n));}////
}}////&#47.0;///////////&#47.h&double theSum(int N){
double An_2;// n&输入项数
if(Nx&0) ///&#47,晕/
////}void main(){
int n=20;叠代
return S;//初值
for(n=0;/数列下一项 分母
S += An_1/An:&/初次没看清题;///&#47,&Nx); //////
An_2=An_1=1.0; ///&#47,An_1,An;
fflush(stdin); ////
printf(&退出方法; An_1=An; ////////, Nx, theSum(Nx));数列之分子分母的特点
S += An_1///
printf(&前%d项的和为:%.15lf&;//////精简版结束 这排版真是怪呀;////
double S;//////
S=0.0; An_2 = An_1 = 1;//小于0项则退出
printf(&附精简版本;///&#47: 输入小于0的数 或用CTRL+C&#92,前导空格一样多时还长短不一, An_1;//
/////////An;
//////// /#include &stdio.h&}void main(){
int Nx;//////////);N;//累加
An_2=An_1;前%d项的和:%.15lf \n&//////前任意项的和,随你便。 完整调试版
double S,An_2;////// n++)
An = An_1+An_2;/&#47
#include&stdio.h&int main(){int f1[21]={0,1,2};int f2[21]={0,2,3};for(i=3;i&21;i++){f1[i]=f1[i-1]+f1[i-2];f2[i]=f2[i-1]+f2[i-2];}//用double精确一些的
for (k = 1; k &=20; k++)//循环前二十项加起来
sum += ((float)f1[i])/((float)f2[i]);//这里多加括号保险一些的
printf(&sum = %.2lf\n&, sum);
return 0;}
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010203040506
03040506070809101112131415161718
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12341281811353117931022241460238113121212221331411423
221156828214542181411231351561248214121313111441522524
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RAW FILE CONVERTER EX powered by SILKYPIX (Ver.3.2.23.0) Installer for Windows 8/ 8.1 / 7 / Vista
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System Requirements
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Supports Direct X 9 or later (recommended)
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.NET Framework 3.5 Service Pack 1 required.
* Other versions of Windows are not supported. Only pre-installed operating s operation is not guaranteed on home-build computers or computers that have been upgraded from earlier versions of Windows.
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The firmware update Ver.3.2.23.0 incorporates the following issues:
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