java 求一个n阶方阵对角线元素之和
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publicclassGetSum{/***@paramargs*/publicstaticvoidmain(String[]args){//TODOAuto-generatedmethodstubi
两条对角线都求了.改变e即改变n#include#include#definee3main(){inti,j,a[e][e];intsum=0;for(i=0;i
importjava.util.Scanner;publicclassTest{publicstaticvoidmain(String[]args){Scannerin=newScanner(Syst
写了一个小时,居然没分啊!算了给你了importjava.util.Scanner;publicclassHelix{/***螺旋输出*/publicstaticvoidmain(String[]ar
#include#defineN10longfun(int(*num)[10],intn){inti,j;longs=1;for(i=0;i再问:能加Q不能另50给你974663046再答:加了,采纳
有的地方修改了下,用动态数组就可以解决.#include#includeintmain(){inti,j=0,sum=0,k=0,n;int**a;printf("请输入行列数:");scanf("%
a11+a22+a33+a31+a13
#include <iostream>using namespace std;void main(){/* 变量定义与初始化
#defineN5intmain(){inti,j,k,jzh[N][N];for(i=0;i
你那个第二题是什么语言的?
#include#include#defineN3voidgetDiagonalValue(inta[N][N]){inti;intsum1,sum2;sum1=sum2=0;for(i=0;i
#include <iostream>#include <iomanip>#include <ctime>using nam
#include#defineN10intgetsum(intn,inta[][N])//要求的通用函数{inti,j,sum=0;for(i=0;i
#include <stdio.h>main(){\x05int a[7][7], i, j, ans;\x05int n;\x
//很简单.采纳吧#include#defineN10longfun(int(*num)[10],intn){inti,j;longs=1;for(i=0;i
假如n等于4,程序如下a=[1234561892111213141516]fori=1:4b(i)=a(i,5-i);endbb'结果为41213
将逆矩阵设出来直接求解请见下图
intsum(inta[][N]){//}
publicclassTest{publicstaticvoidmain(String[]args){double[][]data={{1,2,3},{4,5,6},{7,8,9}};System.o
设n阶方阵:a11,a12,.a1n,a21,a22,.a2n,.,an1,an2,.ann,主对角线和副对角线上的元素之和:(a11+a22+a33+.+ann)+(a1n+a2(n-1)+a3(n