4. 键盘输入3*3的整数矩阵,分别求两条对角线元素的和,使用行指针变量实现.

#includeusingnamespacestd;intmain(){inti,n,s=0;coutn;s=0;for(i=1;i

#includemain(){inti,j,k1,k2,data[3][4];for(i=0;i

#includeintmain(){inta,b,c,t;scanf("%d%d%d",&a,&b,&c);if(a

稍微修改了一下：#includevoidmain(){inta[3][4],i,j,max,b,c;printf("请输入一个3*4的整数矩阵：\n");for(i=0;i再问：为什么第二次还要

#include#includevoidmain(){inti,j,n,number=1,a[30][30];printf("PleaseinputanumberN:");scanf("%d",&n)

#include#defineSIZE5*5intmain(){\x09inta[SIZE],i;\x09for(i=0;i

#includeintmain(){inta,b,c,max;scanf("%d%d%d",&a,&b,&c);max=a;if(b>a)max=b;if(c>a)max=c;printf("max=

#include#definek3intmain(){intenter;//intk=3;intarr[k][k];inti,j,sum=0;for(i=0;i

#includevoidmain(){inta[7][4];intsum=0;inti,j;printf("请输入7*4矩阵：\n");for(i=0;i

自己敲的,你看看是不是你要的……我吃饭去了,有事发邮件479292539@qq.com#include#defineM4#defineN5voidmain(){inta[M][N],i,j;intnu

好久没写过汇编的程序了,代码估计我现在做不出来,能告诉你下原理~把3位十进制分别读入后,存到内存的是相应符号的ASC2码,将这个符号asc码转成数字的asc码比如9存进去的是39H要变成1001,也就

#includevoidmain(){inta,b,c;intm;printf("Pleaseinputabc\n");scanf("%d%d%d",&a,&b,&c);m=a;if(b>m)m=b;

/*#include"stdio.h"main(){intx;printf("请输入一个数字:");scanf("%d",&x);if(x%3==0&&x%5==0)printf("该数字能同时被3,

main函数：Scannerinput=newScanner(System.in);intnum=input.nextInt();if(num%3==0||num%5==0){system.out.p

编程?……_(:з」∠)_再问：恩恩

#include#includeintmain(){inta,sum=0;do{scanf("%d",&a);if((a/100)%10==3)sum+=a;}while(a!=0);printf("

#include#includeintmain(){\x09inta1,a2,a3,max;\x09scanf("%d",&a1);\x09scanf("%d",&a2);\x09scanf("%d"

#include"stdio.h"voidmain(){inta[15];printf("Pleaseinput15numbers:\n");for(inti=0;i

#includeintmain(){inta,b,c;printf("Pleaesinput3int:");scanf("%d%d%d",&a,&b,&c);a=(a>=b)?a:b;a=(a>=c)

#includevoidmain(){inta,b,c;printf("输入三个数：");scanf("%d%d%d",&a,&b,&c);if(a>b&&a>c)printf("最大值为：%d",a