7.编写函数计算斐波那契列F(n) = F(n-1) F(n-2)
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=SUMIF($A$1:$C$5,"张三",$B$1:$D$5)
斐波那契数列至少会给出前2,3项,而从找找规律.这里我们比如是1,2,3,5;则:它的规律是:N1=1,N2=2;N3=N1+N2;N4=N2+N3;...Nn=N(n-2)+N(n-1);int[]
#includeintf(intn){if(n==1)returnn;returnn+f(n-1);}intmain(void){intn;scanf("%d",&n);printf("%d\n",f
用系统内置的函数#include#includevoidmian(){doublepower,f;power=pow(x,y);f=power;printf("f(x,y)=%f",pow(x,y))
帮你写好了.unsigned int fib(unsigned int n) {\x09if (n == 1
#includelongintf(intn){if(n==0)return0;elseif(n==1)return1;elsereturnf(n-1)+f(n-2);}intmain
#includemain(){doublex,y;scanf("%lf",&x);if(x
// C++int F(int n) {if (n == 0) return 1;else if
doublef(doublex){doubles=.0;if(x
第n个元素等于第n-1加n-2个元素调用递归实现啊
#includeintFibonacci(intn){if(n==1||n==2)//递归结束的条件,求前两项return1;elsereturnFibonacci(n-1)+Fibonacci(n-
clearclcx=[50100150200250300350400450 500];y=[4080120160200];z=[0.050.050.050.050.050.250.150.1
#includelongfac(intn){inti;longx=1;for(i=2;i再问:谢谢咯!可是我说的是递归法哦!再答:#includelongfac(intn){if(n==0)retur
=INDIRECT("A1:A"&SUM(B1:B4))
“i=1”---->"i==1","i=2"------>"i==2"
#includeintfibo(intn){if(nreturn1;elsereturnfibo(n-1)+fibo(n-2);}intmain(){intn;scanf("%d",&n);print
这要用到递归算法了,程序如下:#includeintf(intn){if(n==1||n==0)return1;//如果n为0或1,返回1returnf(n-1)+f(n-2);//否则调用自身继续运
cleardimea[20]a[1]=1a[2]=1fori=3to20a[i]=a[i-1]+a[i-2]endforfori=1to30?a[i]if(i%5=0)?endifendfo
用什么语言呢?C还是PASCAL、VB?再问:vc++再答:#include<stdio.h>main(){ longa[30],i; a[0]=1;a[1]=1;&n