统计十个数,统计出正数负数和0的个数编程序

如果觉得好,#includevoidmain(){inta[10],i;intpositive_num,negative_num,zero_num;positive_num=negative_num=

修改如下：#include"stdio.h"intmain(){ints[40],i=0,j,num=0,a=0,b=0,c=0;for(i=0;i

#includeusingnamespacestd;intmain(){inta[20];intsum=0,count1=0,count2=0;inti;for(i=0;i>a[i];if(a[i]=

vari,k,n1,n2:integer;beginn1:=0;n2:=0;fori:=1to20dobeginreadln(k);ifk>0thenn1:=n1+1elseifk

//datafilenameisddd.dat;#include#includemain(){FILE*fp;intt,tag,z,f,l,sz,sf;if((fp=fopen("ddd.dat","

intsign(inta){intb;if(a>0)b=1;elseif(a==0)b=0;elseb=-1;returnb;}main(){intsum[3]={0},x;/*sum[0]为负数的和

统计DA_WORD数据区中正数、0、负数的个数,数据定义如下：DA_WORDDW-1,3,5,0,-5,-7,4,0,-8……COUNTEQU$-DA_WORDNUMDB0；存放正数的个数DB0；存放

#include#includemain(){inti,j,m,n,a[1000];scanf("%d",&i);//数的个数for(j=0;j

using System;using System.Collections.Generic;using System.Linq;using System.Tex

#include#defineRow50//行数最大值voidmain(){inti=0;intj=0;//j为临时变量intarray[Row][101];//二维数组//输入数组do{printf

#include#defineN10intmain(){inti,zh,fu,ling;inta[N];int*pa;pa=a;for(i=0;i

汇编语言DSEGSEGMENTSAVEDB3DUP(?)NUMDW100DUP(?)DSEGENDSCSEGSEGMENTASSUMECS:CSEG,DS:DSEGSTART:MOVAX,DSEGMO

#includeintmain(){inta[10]={1,-1,2,0,5,-6,7,0,-2,0};inti,num_po=0,num_in=0,num_ze=0;for(i=0;i0)num_p

移入累加器A,判断ACC.7

MOVAX,[BX]CMPAX,0JLL1CMPAX,0JGL2CMPAX,0JZL3把ax改成al试试.

#includevoidmain(){inti,a,b,c,t;a=b=c=0;for(i=0;i0)a++;if(t==0)b++;if(t

#includevoidmain(){inta[20];inti;inti1=0;inti2=0;intsum=0;floataver;for(i=0;i

我靠,这么简单的问题你也好意思提,自己动动脑子好不好

将每个数与0相减,差值与一个较小的数（如10E-5）比较,

//C++输入一组非零整数(以输入零作为输入结束标志)到一维数组中,设计程序,求出这组数的平均值,并统计出正数和负数个数#defineMAXNUM1000//数组的最大个数,你可以改#includeu