灵鹊做题网根据整型形参m 计算如下公式的值y＝1 2 1 4 1 6
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/07 06:32:39
y+=j*1.0/(i*i);
#includedoublefun(intm){doubley=1.0;inti;//i改为小写//y+=1.0/(i*i);//1改为1.0return(y);}main(){intn=5;prin
PrivateFunctionFun(mAsInteger)AsSingle'EndFunction
#includevoidmain(){intm,i;floatt=1;printf("请输入m的值：");scanf("%d",&m);for(i=2;i
main函数中的最后一个printf不对应该是printf("theresultism:%lf\n",fun(m));
#include＜conio.h＞#include＜stdio.h＞doublefun(intm){doublet=1.0;inti=2;for(i=2;i＜=m;i++)t+=1.0/(i*(i+1
doublefun(intm);intmain(){inta;scanf("%d",&a);printf("%lf\n",fun(a));}doublefun(intm){doublesum=0.0,
#includeintmain(void){intflag=1,i,n,an;floatresult=0;printf("pleaseinputn:\n");scanf("%n",&n);for(i=
for(intm=1;m
printf("A%d=%f\n",n);输出格式不对改为printf("A%d=%f\n",n,fun(n));后面的逻辑也不正确,给你改了,参考一下.我用的递归（不用递归也能实现）.#includ
π=3.14d=6代入即可：1/4×3.14×6×6=1/4×36×3.14=9×3.14=28.26
#includevoidmain(){inti,c=1,n;floatsum=0;printf("Pleaseinputn:\n");scanf("%d",&n);for(i=1;i
doublefun(intn){doublesum=0.0;for(inti=1,j=1;i
#include"stdio.h"doublesum(intm){\x09inti;\x09doublet;\x09for(i=2,t=1.0;i
doublefun(intm){doublet=1.0;inti;for(i=2;i
不知道是不是你想要的#includedoublefun(intm){inti=1;doublesign=-1;doubles=1;while(i++
#includevoidmain(){intm,i;floata,b,c;a=0;scanf("%d",&m);for(i=2;i
#includeintmain(){intm,i;doubley=0,t=1;scanf("%d",&m);for(i=1;i
double fun(int m){ double t = 1.0;\x09int i&nbs
intfun(intm){inty=1;for(inti=2;i