统计一数组中奇数和偶数的个数分析时间复杂度

8086汇编：设数据存在内存BUF开始的单元中,长度为N,奇数的个数存在ALMOVCX,NMOVAL,0LEADI,BUFL1:TEST[DI],01HJZNEXTINCALNEXT:INCDILOO

//---------------------------------------------------------------------------#includeintmain(void){i

#include#includeusingnamespacestd;intmain(){intxx[200],i=0,cnt1=0,cnt2=0;ifstreamfile;charfilenam

#includeusingnamespacestd;voidmain(){cout

=SUMPRODUCT(((A1:M1-B1:N1)*MOD(COLUMN(A1:M1),2)>0)*1)

连续偶数的次数C19=SUM(N((FREQUENCY(IF(MOD($A$1:$E$17,2)=0,ROW($1:$17)+(COLUMN($A:$E)-1)*17),IF(MOD($A$1:$E$

#includeusingnamespacestd;voidfun2(inta[],intn,int&c1,int&c2){for(inti=0;i

inta[n],b[n];inti=0,j=0for(i=1;i>k;if(kint(k))continue;if(k%2==0){a[i]=k;i++;}else{b[j]=k;j++;}}cout

voidmain(){inta[10]={1,2,3,4,5,6,7,8,9,10};inta=0,b=0,i=0;for(i=0;i

定义20随机数,在判断每个随机数对2求于是否等于0,等于0的话个数加1,这样就好了.

PrivateSubCommand1_Click()Dima(1To20)AsIntegerDimiAsInteger,t1AsInteger,t2AsIntegerFori=1To20a(i)=In

intjs=0;intos=0;int[,]a=newint[4,5];for(inti=0;i

//假定数组为arr[5,4]int even=0;for(int i=0;i<5;i++)   for(int j=0;j<4

#includemain(){inti,a[10],j=0;for(i=0;i

#include <stdio.h>void main(void){ int input[20]; int a=0,b=0;&n

例,求1到100之间偶数和奇数的数量dimiasinteger,aasinteger,basintegerfori=1to100ifimod2=0'偶数a=a+1'偶数奇数else'奇数b=b+1‘奇

privatevoidbutton1_Click(objectsender,EventArgse){int[,]a={{1,2,3,4},{6,7,8,9},{10,11,12,13},{14,15,

#include#include#defineMAX1000intmain(){inta[MAX],i,j,n,p,q,s1,s2;p=0;q=0;s1=0;s2=0;printf("请输入要输入的值

publicclassDemo{publicstaticvoidmain(Stringargs[]){int[]nums={5,2,45,11,13,32,7,24,63,48};intodd=0,e

#include#include#include#includevoidmain(){inti,temp[100],a[50],b[50],c1,c2,b1=0,a1=0;intfun(int);