1. C语言一个3×4的矩阵,输出该矩阵,求矩阵所有元素中的最大值.

#include<stdio.h>#define ROW 3//行数#define COL 4//列数void PlusMatrix(in

#include<stdio.h>#include<stdlib.h>#defineM10#defineN10intmain(void){inti=0,j=0;intArr[M

#includemain(){\x09inta[3][3],i,j,sum=0;\x09printf("请输入数组元素：（3*3）\n");\x09for(i=0;i

#includevoidmain(){inta[3][3];inti,sum1,sum2;for(i=0;i

#include#includeintmain(){intm,n,i,j,s;intarr[30][30];while(scanf("%d%d",&m,&n)!=EOF){for(i=0

//正确的程序如下：#includevoidmain(){inti,j,n,*tem[3];intp[2][3],q[3][2],r[2][2];tem[0]=p[0];tem[1]=q[0];tem

#include<stdio.h>int main(){    int a[2][3];   &n

1:intmain(){intk,n,m=1,a[4][4];for(k=0;k

if(s[i][0]=='0')“'0'”只是一个字符去掉看看.再问：这样我连编译都通不过

#includevoidmain(){\x09inta[3][2]={1,2,3,4,5,6};\x09intb[2][3],i,j;\x09for(i=0;i

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(){

#include#include#include#defineN5voidmain(){\x09intarry[N][N]={0};\x09int\x09i,j,a,b;\x09intn=1;\x09

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

#includeintmain(){intarray[3][3],sum;inti,j;for(i=0;i{for(j=0;j{scanf("%d",&array[i][j]);}}sum=array

#includemain(){inti,j,a[3][4],max;for(i=0;i

这不怎么难啊?再答：#include#defineN3#defineM4voidmain(){inti,j,k,a[N][M];intmax,maxj,flag;printf("Pleaseinput

#includeintmain(){floatst[3][3],sum=0;inti,j;for(i=0;ist[i][j];for(i=0;i

#includevoidmain(){inta[3][4],b[3][4],c[3][4],i,j;printf("pleaseinputaarray:\n");for(i=0;i

#includeintmain(){inta[3][3];inti;intj;for(i=0;i

#includevoidmain(){inta[4][4],max,maxi,maxj,min,mini,minj,i,j;for(i=0;i再问：嗯嗯能解释一下每一步的作用吗⊙﹏⊙再答：-变量定义-