c语言 函数 将一个3*3整形矩阵转值

#includeusingnamespacestd;floata[3][3];intmain(){inti,j;voidchange();cout

用指针转换#includeintmain(void){signedshorta=-1;unsignedshort*b=(unsignedshort*)&a;unsignedshortc=*b;prin

intget_max(inta[4][4]){inti,j;intmax=-0xfffffff;for(i=1;i

合法的整形产量其实就是整形的正确表达,比如十六进制0x0A,八进制书060.十进制数,二进制数,转义字符,字符产量,还有一些其他的表达式,andsoon.y＝（（x=3×y,x+1）,x-1）；是逗号

#include#includeintmain(){inta[3][3]={{1,2,3},{4,5,6},{7,8,9}};intcol[3]={0};//hangintrow[3]={0};//l

if(j==4)\x09\x09\x09\x09printf("%d\n",max);去掉if(j==4)加大括号.改成这样:#defineM3#defineN5#includevoidmain(){

先收藏,等一会在弄.#include#includeintmain(){inti,j,m=0,a[3][3];printf("请输入9个整数\n");for(i=0;i再问：好的，先谢了

#includeintmain(){inta[5][5],i,j,sum;printf("请输入3*3矩阵的元素：\n");for(i=0;i

#includevoidasum(doublea[3][3],doubleb[3][3],doublec[3][3]){inti,j;for(i=0;i

#includeintmain(){inta[10][10],n,i,j;int(*p)[10];voidjzdzz(int(*p)[10],intn);printf("请输入n*n矩阵的阶数

matrix_mul(int**A,int**B,int**C,intm,intp,intn){for(inti=0;i{for(intj=0;j{C[i][j]=0;for(intk=0;k{C[i

#include"stdio.h"intmain(){inti,j,a[3][4];for(i=0;i

#includeintjudge(intx,inty,intz){\x09intm;\x09if(((x+y)>z)&&((x-y)y)&&((x-z)x)&&((y-z)

好吧,简单说下：voidfun(floata[][]){inti,j,floatt;for(i=0;a[i][0];i++)for(j=i;a[i][j];j++){t=a[i][j];a[i][j]

是不是要这样啊.#includeintmain(){inta[3][4]={1,2,3,4,5,6,7,8,9,0,1,2};intb[3][4]={11,21,13,14,51,61,17,18

#includevoidmain(){inti,j;inttemp,res1=0,res2=0;for(i=0;i

#includevoidinput(inta[],intn)//输入{inti;printf("请输入%d个数,用空格隔开\n",n);for(i=0;i

用C语言实现,代码如下#include"stdio.h"#defineM3//矩阵阶数voidmain(){inta[M][M],i,j,sum=0;printf("请输入%d*%d矩阵：\n",M,

intfun(ints){inti=0;while(s!=0){s/=10;i++}returni;}

#include <stdio.h>#define N 3int array[N][N];void transition(int