输出5*5的整数方阵,然后求两条对角线上各元素的和

importjava.io.*;classSEV{publicstaticvoidmain(String[]args)throwsIOException{intmax,min;doublesh;Inp

每个方阵中的长度：2米×5=10米所以总长是：10×5+4×5=70米

#include#include#includevoidavg(inta[],intn,doubleb[]){inti,j;for(i=0;i

inta,b;scanf("%d%d",&a,&b);if(a>100||b>100){if(a>100)printf("%d\n",a);if(b>100)printf("%d\n",b);}els

#include#include#includeintmain(){inta[15],i,j;srand(time(NULL));for(i=0;i<15;i++)a[i]=rand()%100;fo

#includevoidmain(){\x09inta,b,c,min;\x09scanf("%d%d%d",&a,&b,&c);\x09min=a;\x09if(min>b)min=b;\x09if

#includeintmain(){inti,n,sum;printf("请输入整数n:\n");scanf("%d",&n);for(i=1,sum=n;i{printf("%d的%d次方等于%d\

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

给你写了,你看看吧#includeintmain(void){inta[20],i,j,sum=0,temp;for(i=0;i

intSpSum(intv){ints=0;if(v0){s+=mod(v,10);v=v%10;}returnv;}voidEnumNum(intmin,intmax,intx){if(min

请输入:100357111317192329313741434753596167717379838997Pressanykeytocontinue#include#includeintmain(){

方法一：//用数学函数#include#includevoidmain(){inta;scanf("%d",&a);printf("%d\n",abs(a));}方法二：//判断#includevoi

#includevoidfun(intm,intn){\x05printf("%d\n",m*m+n*n);}main(){\x05intm,n;\x05while(scanf("%d%d",&m,&

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("%

#include <stdio.h>#include <stdlib.h>void main(){ int num[5][5]

#include"stdio.h"#include"math.h"main(){inta,b,i,j,k,t,n=0;printf("请输入两个整数:");scanf("%d%d",&a,&b);if

clsrandomizetimerdima(20)fori=1to20a(i)=int(rnd*90+10)printa(i)ifa(i)mod2=0thens=s+1nextiprintsend

1.#includeintmain(void){inti,j;intarr[10],tmp;printf("十个整数：");for(i=0;i{scanf("%d",&arr[i]);

#includeusingnamespacestd;intmain(){intx,y,z;while(1){coutx>>y>>z;cout

#includevoidmain(){inta[6],i,j,k,m;for(i=0;i=0;i--){k=a[5];for(j=4;j>=0;j--)a[j+1]=a[j];a[0]=k;for(m