bool编程输出1~1000中所有个位数是7的素数,统计其个数,并
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#includevoidmain(){inti;for(i=1;i
//提供一个方法片段,供参考!Eclipse通过的哦.//和=500500;i0=32.publicvoidshow0(){inti=1,i0=-1,sum=0;while(i500)&&(i0==-
#include"stdio.h"voidmain(){inti;for(i=1;i
#includeintmain(){intn;n=0;for(inti=1;i再问:大侠,这是你自己做的吗?有没有把握对啊?再答:是,刚写的,不信你运行一下。
#include#defineN4main(){inti,j,a[N][N]={{3,5,6,3},{0,8,9,1},{0,0,5,0},{0,0,0,7}};for(i=0;ifor(j=0;ji
importjava.util.Random;publicclassTest1{publicstaticvoidmain(String[]args){Randomr=newRandom();intju
#includevoidmain(){intn,i,j=0,a[1000];for(i=1;i
PrivateSubCommand1_Click()Dimi,j,a,b,sFori=7To997a=FalseForj=1Toi-1IfiModj=0Andj>1Thena=TrueExitForE
privatesubform_click()dimnaslongdimxasstringdimyasstringfori=1000to9999x=mid(i,1,2)y=mid(i,4,1)&mid(
voidmain(){inti=0,num=0;for(i=1;i{if((i%10==3)&&(i%3!=0)){printf("%3d",i);num++;}if(num==10){num=0;p
#includevoidmain(){inti;for(i=1;i
#includevoidmain(){inti;for(i=1;i
if(a>b)k=true;elsek=false;
sum=0fori=1to100if(imod3)thensum=sum+inextiprintsum
PrivateSubForm_Load()ForI=1To1000IfIMod7=0ThenX=X+1IfXMod10=0ThenLabel1.Caption=Label1.Caption&vbNew
#includemain(){intarray[3][4];inti,j,max;for(i=0;i
clear?"1~999能被3整除且至少有一位是5的数有:"fori=1to999ifi%3=0and"5"$alltrim(str(i))?iendifendfo
IconfessthatI'mansweringfortask.#includeintmain(){intn,m=1,i=1;scanf("%d",&n);if(n{printf("error\n")
settalkoffcleaa=1b=1c=1dowhilc
#include#includeintmain(intargc,char**argv){intnum[1000];intmin=10000;inti;for(i=0;i再问:找到问题了--那是i++不