5*5矩阵对角线元素之和-搜答案-灵鹊做题网

#include#includeconstintMAX=100000;intgcd(inta,intb){returna%b?gcd(b,a%b):b;}booldig(charx){returnx>

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

不是指一个矩阵化简之后的矩阵；111205243这个矩阵的主对角线上的元素是1、0、3

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

#includeusingnamespacestd;voidmain(){intCArray[5][5];inti,j;intsum=0;cout

#includeintmain(){inta[5][5];for(inti=0;i<5;i++){for(intj=0;j<5;j++)scanf("%d",&a[i][j]);}ints=0;for

#includeusingnamespacestd;intmain(){inta[5][5];inti,j,s1=0,s2=0;cout

#include#defineN6main(){inti,j,n=1,s=0,m=0,a[N][N];for(i=0;i

'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

#includeintmain(){\x09inta[4][4],i,j,msum=0,ssum=0;\x09for(i=0;i\x09\x09for(j=0;j\x09\x09\x09printf(

#include <stdio.h>void main(){\x05int i, j, array[5][5], sum=0;\

//zd_40.cpp:Definestheentrypointfortheconsoleapplication.//#includeintmain(intargc,char*argv[]){inti

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

可以计算任意矩阵的对角线,把N改了就是：#defineN3main(){inti,j,a[N][N];intsum=0;printf("\npleaseinputthearray:\n");for(i