实现功能:计算正整数m到n之间所有完数之和.-搜答案-百度作业网

#include#include#include/*利用辗转相除法求最大公约数*/intgcd(intn,intm){intr;if(n

改一下输出格式控制符就可以了printf("%f",a);改为printf("%g",a);改过的程序：#includedoubledeal(doublex,doubleh){doublea;inti

main(){intm,n,k,z;scanf("%d,%d",&m,&n);for(k=0;k>0;)scanf("%d",k);结果精确到小数点后k位,貌似不太可能,好象声明时要用常数z=m/n;

这个算起来属于大整数除法.因为整数大小是不确定的,而且正整数k可能也比较,用float,double根本就解决不了问题.

证明：令m/n=t(t>=0)则m=nt(m+7*n)/(m+n)=(t+7)*n/n(t+1)n不为零原式=（t+7）/(t+1)=1+6/(t+1)1)0根号7则1+6/(t+1)

然后是用递归逆转输出数字的.programReverseNum;varnum:integer;procedurefun(n:integer);beginifn1thenbeginwrite(nmod1

#include <stdio.h>void main(){int i;for(i=100;i<=300;i++){if(i%3==0&

#include<stdio.h>void gcd(int m, int n) {int temp = m

#include<stdio.h>int gcd(int m,int n){while (n){int t=m%n;m=n;

#include#includeintprime(inti){intk;for(k=2;k

double fun(int n){\x09int i,j;\x09double s=0;\x09for(i=3;i<=n;i++)\x09{\x09\x

#includelongfun(longa);voidmain(){longm,n;printf("pleaseinputthe'm'\n");scanf("%ld",&m);printf("plea

#include <stdio.h>void findNum (int, int);int main () {&nbs

#include <iostream>using namespace std;int main(){

设这六个连续完全平方数中最小的为K²,最大的为(K+5)²,则根据题意,应满足：(K+5)²-K²（因为如果有：(K+6)²-K²上面两式解

#include#includeintmain(intargc,char*argv[]){intm,n,c,i,j,max;scanf("%d%d",&m,&n);if(m=2){printf("Th

#includeintfact(intn){intt=1;for(inti=1;i

用什么语言?再问：c��再答：#include#defineN5main(){intn;intfun();for(n=1;n

#includeusingnamespacestd;intmain(){inta,b;couta>>b;inttemp,num1,num2;if(a{temp=a;a=b;b=temp;}num1=a

两种可能当m是偶数时,它们相等当m是奇数时,它们互为相反数