统计NUM开始的10个数中正数.负数.0的个数,并显示输出

#includevoidmain(){inti,count=0;floata[10],sum=0,aver;printf("pleaseinput10number:\n");for(i=0;i

=COUNTIF(A1:A100,">0")

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

packagetest;importjava.util.Random;publicclassrandomNum{publicstaticvoidmain(String[]args){chec

假设这些数字写在A列则公式为=COUNTIF(A:A,">=650")-COUNTIF(A:A,">750")如果是07及以上的版本公式可以写成=COUNTIFS(A:A,">=650",A:A,"再

DATASEGMENTBUFDW12,0,-13,45,-23,0,-34,-12,34ZHENGDB0FUDB0COUNTEQU(ZHENG-BUF)/2DATAENDSCODESEGMENTASS

貌似以前是单片机里面的题目很简单.但是现在不记得了.MOVR0,#60HMOVX@R0,30HMOVDPTR,#60HMOVXA,@DPTRMOVDPTR,#2000HMOVX@DPTR,A请参考

//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

环数相加除以射击次数就是你的平均环数,望采纳.

#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

publicstaticvoidtest(){Scannersc=newScanner(System.in);longnum=0,negative=0,positive=0,sum=0;Listnum

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

1.clearstore0ton1,n1dowhile.t.input"输入一个正数或负数："toxifx>0n1=n1+1elsen2=n2+1endifwait"还要继续输入吗?(Y/N)"toy

intb;inti;intnum[3]={0,0,0};intcount[];int*p=count;scanf("%d",&b);for(i=0;i再答：#includevoidmain(){int

我只看见40个啊