编写函数fun,求一分数序列2 1
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intfun(intn){inta=n,b=0;while(a>0){b=b*10;b=b+a%10;a=a/10;}printf("%d",b);getch();return0;}或者把后三行删掉,
能给出分数序列的规律,貌似是前项的分子与分母的和为下一项的分子,前一项的分子为下一项的分母啊.然后再变成就很简单了,用for循环
floatfun(intn){\x09inta[30];\x09floatsum=2/1;\x09a[0]=2;\x09a[1]=3;\x09for(inti=2;i再问:floati;doublea
你需要一个分数式的结果?再问:什么结果都可以再答:#include"stdio.h"#include"math.h"voidPrintInfo();intmain(){doublesum,buf;in
intfun(intn){intm=1,sum=0,i,j;for(i=1;i
F(n)=(1/2)*((sqrt(5)+1)^(n+2)-(1-sqrt(5))^(n+2))/((sqrt(5)+1)^(n+1)-(1-sqrt(5))^(n+1))
Private Sub Command1_Click()Dim a As Long, b As Long, c
#include"stdio.h"intmain(){inta1=1,a2=2,n=0,i=1,tmp=a2;doublesum=0.0;printf("输入n:");scanf("%d",&n);w
voidfun(longn){intwan,qian,bai,shi,ge;//定义各数位longnixushu;//定义逆序数wan=n/10000;qian=(n-wan*10000)/1000;
intfun(intn){inta=n,b=0;while(a>=1){b=10*b+a%10;a=a/10;}returnb;}已经调试通过了哦!
PrivateFunctionSumA(ByValaAsInteger)AsLongDimiAsInteger,NAsIntegerFori=500To600IfiModa=0ThenN=N+iNex
#include#includevoidswap(intc[],intlen){inti=0;inttmp;for(;i{tmp=c[i];c[i]=c[len];c[len]=tmp;}}intmu
#includevoidmain(){inti,t,n=20;floata=2,b=1,s=0;for(i=1;i
请楼主参考采纳intfun(intt){inti;intcurrent=0;for(i=1;totalt)break;current+=i;}returncurrent;}
Functions(ByValaAsInteger,ByValbAsInteger)AsIntegerDimiAsIntegerFori=1To1000IfiModa=0AndiModb=0Thens
1.intf1=0,f2=1,f3;2.returnf3;继续写:f3=f1+f2;f2=f3;f1=f2;
#includeusingnamespacestd;longunsignedfun(intn){if(n>1)returnn*fun(n-1);return1;}voidmain(){intn;cou
functionfun(d,h){if(d
intfun(w){intsum=0;for(inti=1;i