求一个5*5整形矩阵对角线元素之和-搜答案-搜狗做题网

#include"stdio.h"#defineM5//矩阵阶数voidmain(){inta[M][M],i,j,sum=0;printf("请输入%d*%d矩阵：\n",M,M);for(i=0;

for(i=0;i=0;i--,j++)sum1=sum1+a[i][j];

#include#includevoidmain(){inta[3][3];inti,j;//i为行j为列intsum=0;intsum1=0;//输入for(i=0;i

#includevoidmain(){inti,j,a[3][3],sum=0;for(i=0;ifor(j=0;jscanf("%d",&a[i][j]);for(i=0;isum+=a[i][i]

publicclassGetSum{/***@paramargs*/publicstaticvoidmain(String[]args){//TODOAuto-generatedmethodstubi

intnSum=0;for(inti=0;i

1+5+9+7+3

晕.main(){inti,j;inta[3][3];intresult=0;for(i=0;i

intmatrix(int**aintra,intca){intsum=0;for(inti=0;i再问：��д�� [][]�

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

对角线元素之和可以调用库函数trace,也就是矩阵的迹定义了矩阵A后,直接调用函数trace(A).

intmain(){inta[5][5],i,j,sum1,sum2;sum1=0;sum2=0;for(i=0;i

PrivateSubCommand1_Click()DimmArr(1To5,1To5),r%,c%,tmp%Forr=1To5Forc=1To5Randomizetmp=Int(Rnd*90)+10

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

#includevoidmain(){inta[3][3],i,j,sum=0;cout

'定义事件由单击按钮触发providesubcommand1_click()'定义数组dima(1to5,1to5)asinteger'生成矩阵数组元素fori=1to5forj=1to5a(i,j)

#includeintmain(void){inti,j,sum;inta[5][5];sum=1;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,

voidmain(){inta[4][4]={1,2,3,4,5,6,7,8,9,0,11,12,13,14,15,16};intmin,max;inti,j;min=max=a[0][0];for(

可参考以下程序:inta[5][5];//5*5数组inti,x,y;//x,y是两个对角线元素的和.x=0;for(i=0;i