形成一个5*5矩阵,要求对角线元素均为1,其余元素为零-搜答案-灵鹊做题网

#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];

intnSum=0;for(inti=0;i

#include"stdio.h"voidmain(){intmagic[5][5]={{17,24,1,8,15},{23,5,7,14,16},{4,6,13,20,22},{10,12,19,

eye([3,5])主对角线全1ans=100000100000100fliplr(eye([3,5]))副对角线全1ans=000010001000100

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

#include<cstdlib>#include<ctime>using namespace std;int main(){ sran

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

for(sum=0,i=0;i

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

最直观的办法,直接构造数组doublematrix[5][5]{{1,0,0,0,1},{0,1,0,1,0},{0,0,1,0,0},{0,1,0,1,0},{1,0,0,0,1},};

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

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

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

DimA(5,5)AsInteger,iAsInteger,jAsInteger,sAsStringFori=1To5Forj=1To5Randomize'加上这句,用于修改生成随机数的种子,否则总是

#includeintmain(void){inti,j,sum;inta[5][5];sum=1;for(i=0;i

在matlab中输入x=[0111010101110111010101110]然后回车(注意要用英文输入法,我临时也就会这么笨的方法,幸好你的矩阵也不大)

#include"stdio.h"#defineN10intmian(){inta[N][N];inti,j,sum=0;for(i=0;i

//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