从键盘上输入a与n的值,计算a aa aaa aaaa -(共n项)的和
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publiclongpower(intm,intn){if(nreturnm;elsereturnpower(m,n--)*m;}
#include <stdio.h>void main(){int i,j,a,n,k=0,out=0;printf("请输入a与n:")
#includeintGetNumber(intn)//用递归来实现很简单{intsum=0;if(n/10!=0){\x09sum+=GetNumber(n/10);}sum+=n%10;retur
importjava.util.Scanner;publicclassA{//下面的方法是对单个数求阶乘,并返回publicstaticintjieCheng(intn){intsum=1;for(i
先使用读入函数,从键盘读入三个数.你这三个数分别是一元二次方程的ax^2+bx+c=0中的abc,那么你就是用公式先判断△=b^2-4ac的情况,分三种1△≥0有两个实数根x=[-b±(b^2-4ac
#includedoublearea(doublea,doubleb,doubleh){return0.5*(a+b)*h;//二分之一上底加下底的和乘以高}intmain(void){doublea
我给你写了一个,结果保留了4位有效数字#include<iostream>#include<cstdio>using namespace std;int&n
#include#includelongfac(intn,inta){longsum;if(n==1){sum=a;}else{sum=(long)(pow(10,n-1)*a)+fac(n-1,a)
#includevoidmain(void){intn=0;inti=0;floats=0;floatfTemp=0.0f;scanf("%d",&n);for(i=1;i
#include<stdio.h>#define N 10int main(){ int i,*max
#include#includevoidmain(){doublex;doubley;coutx;if(-2
#includeintmain(){chara,b;a=getchar();scanf("%c",&b);printf("a=%c\n",a);printf("b=%c\n",b);return0;}
#include<stdio.h>void main( ){ int&nb
#include#includemain(){\x09floata,b,c,s,area;\x09scanf("%f%f%f",&a,&b,&c);\x09s=(a+b+c)/2;\x09if(a>b
VC6.0调试通过了,while循环结构#include"stdio.h"voidmain(){inti,sum=0,n;i=1;printf("Inputn:");scanf("%d",&n);wh
ints=0;for(inti=0;i
#includevoidmain(){inta,b,sum;scanf("%d,%d",&a,&b);sum=a+b;printf("thesumis%d",sum);}
#include#includeintmain(){inttemp;inta,b,c;scanf("%d%d%d",&a,&b,&c);if(a>b)//保证a中存放最大值{temp=b;b=a;a=
#include<stdio.h>int fx(int x,int y){int sum=0,a=x;;for(int i=0;i&l
#include<stdio.h>int main(void){int i,j,a,n,sum,total=0;scanf("%d%d",&am