给定一系列正整数，请按要求对数字进行分类，并输出以下5个数字：
A1 = 能被5整除的数字中所有偶数的和；
A2 = 将被5除后余1的数字按给出顺序进行交错求和，即计算n1-n2+n3-n4...；
A3 = 被5除后余2的数字的个数；
A4 = 被5除后余3的数字的平均数，精确到小数点后1位；
A5 = 被5除后余4的数字中最大数字。
[编程题]数字分类 (20)
- 热度指数：110222 时间限制：C/C++ 1秒，其他语言2秒 空间限制：C/C++ 32M，其他语言64M
- 算法知识视频讲解
输入描述:
每个输入包含1个测试用例。 每个测试用例先输入一个不超过1000的正整数N。 然后给出N个不超过1000的待分类的正整数。数字间以空格分隔。
输出描述:
对给定的N个正整数，按题目要求计算A1~A5并在一行中顺序输出。数字间以空格分隔，但行末不得有多余空格。
若其中某一类数字不存在，则在相应位置输出“N”。
示例1
输入
13 1 2 3 4 5 6 7 8 9 10 20 16 18
输出
30 11 2 9.7 9
说明
总共13个数字
不难就是费事，淦
import java.math.BigDecimal;
import java.math.RoundingMode;
import java.text.DecimalFormat;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Mian c = new Mian();
Scanner in = new Scanner(System.in);
System.out.println("请输入数的个数");
int N = in.nextInt();
int[] arr = new int[N];
for (int i = 0; i < N; i++) {
arr[i] = in.nextInt();
}
String[] s = new String[5];
if(c.Even_Sum(arr,N) ==-1)
s[0] ="N";
else {
s[0] = String.valueOf(c.Even_Sum(arr,N));
}
if(c.Chu5yu1(arr,N) == -1)
s[1] ="N";
else {
s[1] = String.valueOf(c.Chu5yu1(arr,N));
}
if(c.Yu2Num(arr,N) == -1)
s[2] ="N";
else {
s[2] = String.valueOf(c.Yu2Num(arr,N));
}
if(c.Aver(arr,N) == -1)
s[3] ="N";
else {
s[3] = String.valueOf(c.Aver(arr,N));
}
if(c.Find_Max(arr,N) == -1)
s[4] ="N";
else {
s[4] = String.valueOf(c.Find_Max(arr,N));
}
for(String s2: s){
System.out.print(s2+" ");
}
}
public int Even_Sum(int[] aarr, int N) {
int sum = 0;
int count = 0;
for (int i = 0; i < N; i++) {
if (aarr[i] % 5 == 0 && aarr[i] % 2 == 0) {
sum += aarr[i];
count++;
}
}
if (count > 0)
return sum;
else
return -1;
}
public int Chu5yu1(int[] aarr, int N) {
int sum = 0;
int count = 0;
int count1 = 0;
for (int i = 0; i < N; i++) {
if (aarr[i] % 5 == 1) {
if (count1 % 2 == 0) {
sum += aarr[i];
} else {
sum -= aarr[i];
}
count1++;
}
}
if (count1 > 0)
return sum;
else return -1;
}
public int Yu2Num(int[] arr, int N) {
int count = 0;
for (int i = 0; i < N; i++) {
if (arr[i] % 5 == 2) {
count++;
}
}
if (count > 0)
return count;
else return -1;
}
public double Aver(int[] arr, int N) {
double count = 0;
double Sum = 0;
double sum = 0;
for (int i = 0; i < N; i++) {
if (arr[i] % 5 == 3) {
sum += arr[i];
count++;
}
}
if (count > 0) {
Sum = sum / count;
// DecimalFormat df = new DecimalFormat("#.00");
// String formattedvalue = df.format(Sum);//字符串形式
BigDecimal bd = BigDecimal.valueOf(Sum).setScale(1, RoundingMode.HALF_UP);
return bd.doubleValue();
} else return -1;
}
public int Find_Max(int[] arr, int N) {
int count = 0;
int max = 0;
for (int i = 0; i < N; i++) {
if (arr[i] % 5 == 4) {
count++;
if (arr[i] > max) max = arr[i];
}
}
if (count > 0)
return max;
else return -1;
}
}
发表于 2025-01-06 11:58:36
回复(0)
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int a1 = 0;
int a2 = 0;
int a3 = 0;
double a4 = 0;
int a5 = 0;
int i = 0;
int temp = 1;
int count = 0;
while (i++ < n){
int value = scanner.nextInt();
if (value % 5 == 0 && value % 2 == 0) a1 += value;
if (value % 5 == 1){
a2 += temp * value;
temp *= -1;
}
if (value % 5 == 2) a3++;
if (value % 5 == 3) {
a4 += value;
count++;
}
if (value % 5 == 4 && value > a5) a5 = value;
}
if (a1 != 0) System.out.print(a1 + " ");
else System.out.print("N ");
if (a2 != 0) System.out.print(a2 + " ");
else System.out.print("N ");
if (a3 != 0) System.out.print(a3 + " ");
else System.out.print("N ");
if (a4 != 0) System.out.format("%.1f ",a4/count);
else System.out.print("N ");
if (a5 != 0) System.out.print(a5);
else System.out.print("N");
scanner.close();
}
}
发表于 2022-03-03 14:46:15
回复(0)
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner scan=new Scanner(System.in);
int N=scan.nextInt();
int a1=0;
int a2=0;
boolean b2=false;
boolean add=true;
int a3=0;
double a4=0;
int num4=0;
int a5=0;
for (int i = 0; i < N; i++) {
int x=scan.nextInt();
int m=x%5;
switch(m){
case 0: if(x%2==0) a1+=x;break;
case 1: {
b2=true;
if(add){
a2+=x;
}else{
a2-=x;
}
add=!add;
break;
}
case 2: a3++;break;
case 3: {
num4++;
a4+=x;
break;
}
case 4: {
if(x>a5) a5=x;
break;
}
}
}
String A1=(a1==0)?"N":Integer.toString(a1);
String A2=(b2)?Integer.toString(a2):"N";
//a3
String A3=(a3==0)?"N":Integer.toString(a3);
String A4;
if(num4==0){
A4="N";
}else{
a4=Math.round(10*a4/num4)/10.0;
A4=Double.toString(a4);
}
String A5=(a5==0)?"N":Integer.toString(a5);
System.out.printf("%s %s %s %s %s%n",A1,A2,A3,A4,A5);
}
}
发表于 2020-03-08 15:48:49
回复(0)
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
int A1=0;
int A2=0;
int A3=0;
float A4=0;
int A5=0;
float s=0;
int q=0;
int count=1;
String str1 = null;
Scanner sc =new Scanner (System.in);
int sum=sc.nextInt();
int max=Integer.MIN_VALUE;
int [] a =new int [sum];
for(int i=0;i<a.length;i++) {
a[i]=sc.nextInt();
}
for(int i=0;i<a.length;i++) {
/*
* A1 = 能被5整除的数字中所有偶数的和；
*/
if(a[i]%5==0&&a[i]%2==0) {
A1+=a[i];
}
/*
* A2 = 将被5除后余1的数字按给出顺序进行交错求和，即计算n1-n2+n3-n4...；
*/
if(a[i]%5==1) {
if(count%2==1)
a[i]=a[i];
else
a[i]=(-1)*a[i];
A2=A2+a[i];
count++;
}
/*
* A3 = 被5除后余2的数字的个数；
*/
if(a[i]%5==2) {
A3+=1;
}
/*
* A4 = 被5除后余3的数字的平均数，精确到小数点后1位；
*/
if(a[i]%5==3) {
// System.out.println(a[i]);
s+=a[i];
q++;
A4=s/q;
str1 = String.format("%.1f",A4);
}
/*
* A5 = 被5除后余4的数字中最大数字。
*/
if(a[i]%5==4) {
if(a[i]>max) {
max=a[i];
}
A5=max;
}
}
if(A1==0) {
System.out.print("N"+" ");
}else {
System.out.print(A1+" ");
}
if(A2==0) {
System.out.print("N"+" ");
}else {
System.out.print(A2+" ");
}
if(A3==0) {
System.out.print("N"+" ");
}else {
System.out.print(A3+" ");
}
if(str1==null) {
System.out.print("N"+" ");
}else {
System.out.print(str1+" ");
}
if(A5==0) {
System.out.print("N");
}else {
System.out.print(A5);
}
}
}
public class Main{
public static void main(String[] args) {
int A1=0;
int A2=0;
int A3=0;
float A4=0;
int A5=0;
float s=0;
int q=0;
int count=1;
String str1 = null;
Scanner sc =new Scanner (System.in);
int sum=sc.nextInt();
int max=Integer.MIN_VALUE;
int [] a =new int [sum];
for(int i=0;i<a.length;i++) {
a[i]=sc.nextInt();
}
for(int i=0;i<a.length;i++) {
/*
* A1 = 能被5整除的数字中所有偶数的和；
*/
if(a[i]%5==0&&a[i]%2==0) {
A1+=a[i];
}
/*
* A2 = 将被5除后余1的数字按给出顺序进行交错求和，即计算n1-n2+n3-n4...；
*/
if(a[i]%5==1) {
if(count%2==1)
a[i]=a[i];
else
a[i]=(-1)*a[i];
A2=A2+a[i];
count++;
}
/*
* A3 = 被5除后余2的数字的个数；
*/
if(a[i]%5==2) {
A3+=1;
}
/*
* A4 = 被5除后余3的数字的平均数，精确到小数点后1位；
*/
if(a[i]%5==3) {
// System.out.println(a[i]);
s+=a[i];
q++;
A4=s/q;
str1 = String.format("%.1f",A4);
}
/*
* A5 = 被5除后余4的数字中最大数字。
*/
if(a[i]%5==4) {
if(a[i]>max) {
max=a[i];
}
A5=max;
}
}
if(A1==0) {
System.out.print("N"+" ");
}else {
System.out.print(A1+" ");
}
if(A2==0) {
System.out.print("N"+" ");
}else {
System.out.print(A2+" ");
}
if(A3==0) {
System.out.print("N"+" ");
}else {
System.out.print(A3+" ");
}
if(str1==null) {
System.out.print("N"+" ");
}else {
System.out.print(str1+" ");
}
if(A5==0) {
System.out.print("N");
}else {
System.out.print(A5);
}
}
}
发表于 2019-07-25 00:55:14
回复(0)
import java.text.DecimalFormat;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Main{
public static void main(String[] args){
int i = 0;
Scanner input = new Scanner(System.in);
int N = input.nextInt();
int[] nums = new int[N];
while(i <N){
nums[i++] = input.nextInt();
}
NumberSorting(nums);
}
private static void NumberSorting(int[] nums){
boolean flag1 = false,flag2 = false,flag3 = false,
flag4 = false,flag5 = false;
int A1Sum = 0;
int A2Sum;
int A3Count = 0;
String A4Average;
int maxnum = 0;
List<Integer> lst = new ArrayList<>();
List<Integer> lst1 = new ArrayList<>();
List<Integer> lst2 = new ArrayList<>();
for (int num : nums) {
if (num % 5 == 0) {
lst.add(num);
}
if (num % 5 == 1) {
lst1.add(num);
flag2 = true;
}
if (num % 5 == 2) {
A3Count++;
flag3 = true;
}
if (num % 5 == 3) {
lst2.add(num);
flag4 = true;
}
if (num % 5 == 4) {
maxnum = Math.max(num, maxnum);
flag5 = true;
}
}
for (Integer val : lst) {
if (val % 2 == 0) {
A1Sum += val;
flag1 = true;
}
}
int sum = 0,sum1 = 0;
for (int i = 0; i <lst1.size() ; i++) {
if(i %2 == 0){
sum += lst1.get(i);
}else{
sum1 += lst1.get(i);
}
}
A2Sum = sum - sum1;
int sum2 = 0;
for(Integer e : lst2){
sum2 += e;
}
DecimalFormat df = new DecimalFormat("#.0");
A4Average = df.format((sum2 / (double)lst2.size()));
if(flag1){
System.out.print(A1Sum+" ");
}else{
System.out.print("N"+" ");
}
if(flag2){
System.out.print(A2Sum + " ");
}else {
System.out.print("N"+" ");
}
if(flag3){
System.out.print(A3Count+" ");
}else{
System.out.print("N"+" ");
}
if(flag4){
System.out.print(A4Average+" ");
}else {
System.out.print("N"+" ");
}
if(flag5){
System.out.print(maxnum);
}else {
System.out.print("N");
}
}
}
发表于 2019-06-13 22:25:40
回复(0)
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner in = new Scanner(System.in); int i = in.nextInt(); int flag = 0; int A1 = 0, A2 = 0, A3 = 0, A4 = 0, A4_sum = 0, A5 = 0; while (i > 0) { int anInt = in.nextInt(); int yuShu = anInt % 5; if (yuShu == 0) { if (anInt % 2 == 0) { A1 += anInt; } } else if (yuShu == 1) { A2 += Math.pow(-1, flag) * anInt; flag++; } else if (yuShu == 2) { A3++; } else if (yuShu == 3) { A4_sum += anInt; A4++; } else { if (anInt>A5){ A5 = anInt; } } i--; } String A1_str = "N"; String A2_str = "N"; String A3_str = "N"; String A4_str = "N"; String A5_str = "N"; if (A1 != 0) { A1_str = String.valueOf(A1); } if (A2 != 0) { A2_str = String.valueOf(A2); } if (A3 != 0) { A3_str = String.valueOf(A3); } if (A4 != 0) { A4_str = String.format("%.1f", A4_sum / (double) A4); } if (A5 != 0) { A5_str = String.valueOf(A5); } System.out.println(A1_str + " " + A2_str + " " + A3_str + " " + A4_str + " " + A5_str); } }
编辑于 2019-05-15 08:33:43
回复(0)
import java.util.Scanner;
public class Main{
public static void main(String [] args){
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
int N = sc.nextInt();
int[] input = new int[N];
for(int i=0; i<N; i++){
input[i] = sc.nextInt();
}
int[] a = new int[5];
int max = 0;
int count = 0; // 记录平均数要算几个
int seq = 1; // 交错加减时要用
for(int i=0; i<N; i++){
int c = input[i]%5;
if(c == 0){
if(input[i]%2 == 0){
a[0]+=input[i];
}
}else if(c == 1){
if(seq%2 == 1){
a[1]+=input[i];
}
else{
a[1] -= input[i];
}
seq++;
}else if(c == 2){
a[2]++;
}else if(c == 3){
a[3]+=input[i];
count++;
}else {
if(input[i] > a[4]){
a[4] = input[i];
}
}
}
for(int i=0; i<4; i++){
if(a[i] == 0){
System.out.print("N ");
}else if(i == 3){
double res = (double)a[i]/count;
System.out.print(String.format("%.1f",res) + " ");
}else{
System.out.print(a[i] + " ");
}
}
if(a[4] == 0){
System.out.print("N");
}else {
System.out.print(a[4]);
}
}
}
}
发表于 2019-03-17 09:53:26
回复(0)
import java.util.Scanner;
public class Main{
public static void main(String[] args) {
Scanner in=new Scanner(System.in);
while(in.hasNext()) {
int n=in.nextInt();
int []num=new int[n];
for(int i=0;i<num.length;i++) {
num[i]=in.nextInt();
}
int A1=0;int a=0;
int A2=0;int b1=0;
int A3=0;int c=0;
int A4=0;
int A5=0;int e=0;
int b=1;double d=0;
for(int i=0;i<num.length;i++) {
if(num[i]%5==0&&num[i]%2==0) {
A1+=num[i];a++;
}
if(num[i]%5==1) {
A2=A2+b*num[i];
b=-b;b1++;
}
if(num[i]%5==2) {
A3++;c++;
}
if(num[i]%5==3) {
d++;
A4+=num[i];
}
if(num[i]%5==4) {
if(num[i]>A5) {
A5=num[i];e++;
}
}
}
//System.out.print(a);
if(a==0) {
System.out.print("N ");
}else {
System.out.print(A1+" ");
}
System.out.print(b1!=0 ? A2+" ":"N ");
System.out.print(c!=0 ? A3+" ":"N ");
System.out.print(d!=0 ? String.format("%.1f", A4/d)+" ":"N ");
System.out.print(e!=0 ? A5:"N ");
}
}
}
发表于 2019-03-06 11:22:30
回复(0)
什么鬼
import java.util.Scanner;
import java.text.DecimalFormat;
public class Main{
public static void main(String[] args){
Scanner scanner = new Scanner(System.in);
Integer num_A1 = 0;
Integer num_A2 = 0;
Integer num_A3 = 0;
Integer num_A4 = 0;
Integer num_A5 = 0;
int num_A4_count = 0;
int lol = -1;
boolean b_A1 = false;
boolean b_A2 = false;
boolean b_A3 = false;
boolean b_A4 = false;
boolean b_A5 = false;
while(scanner.hasNext()){
Integer tmp = scanner.nextInt();
switch(tmp%5){
case 0:
if(tmp%2 ==0){
b_A1 = true;
num_A1 += tmp;
}
break;
case 1:
b_A2 = true;
lol = (-1)*lol;
num_A2 += lol*tmp;
break;
case 2:
b_A3 = true;
num_A3++;
break;
case 3:
b_A4 = true;
num_A4 += tmp;
num_A4_count++;
break;
case 4:
b_A5 = true;
if(tmp>num_A5){
num_A5 = tmp;
}
break;
}
}
System.out.println((b_A1?num_A1:"N")+" "+(b_A2?num_A2:"N")+" "+(b_A3?num_A3:"N")+" "+(b_A4?(new DecimalFormat("#.0").format(num_A4/num_A4_count)):"N")+" "+(b_A5?num_A5:"N"));
}
}
发表于 2018-09-20 16:21:57
回复(0)
我的小数点后一位格式化用的方法稍有不一样(不是最佳的)
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
numberClassify();
}
static void numberClassify() {
Scanner scanner = new Scanner(System.in);
int num = scanner.nextInt();
int A1 = 0, A2 = 0, A3 = 0, A5 = 0;
int factor = -1;
int numA4 = 0;
float A4 = 0;
for (int i = 0; i < num; i++) {
int number = scanner.nextInt();
switch (number % 5) {
case 0:
if (number % 2 == 0) {
A1 += number;
}
break;
case 1:
factor = -factor;
A2 += factor * number;
break;
case 2:
A3++;
break;
case 3:
numA4++;
A4 += number;
break;
case 4:
if (number > A5) {
A5 = number;
}
break;
default:
}
}
// System.out.println(A4);
if (numA4 != 0) {
int temp = Math.round(A4 * 10 / numA4);
A4 = (float) temp / 10;
}
System.out.println((A1 == 0 ? "N" : A1) + " "
+ (A2 == 0 ? "N" : A2) + " "
+ (A3 == 0 ? "N" : A3) + " "
+ (numA4 == 0 ? "N" : A4) + " "
+ (A5 == 0 ? "N" : A5));
}
}
采用这个
new DecimalFormat("#.0").format(A4 / numA4))
或许会更好。
发表于 2018-06-04 20:28:24
回复(0)
importjava.text.DecimalFormat;//数字格式化包
importjava.util.*;
publicclassMain {
publicstaticvoidmain(String args[]) {
Scanner sc=newScanner(System.in);
inta=0,b=0,c=0,max=0,q=1,s=0;
DecimalFormat df=newDecimalFormat("#.0");//创建格式化对象,格式为保留小数点后1位
doubled=0.0;
intN=sc.nextInt();//输入N
intarr[]=newint[N];//创建数组容纳数据
for(inti=0;i<arr.length;i++) {
arr[i]=sc.nextInt();//开始写入数据
}
for(intx:arr) {//开始遍历数组
if(x%5==0&&x%2==0) {//A1，能被5整除的偶数和
a=x+a;
}
if(x%5==1) {//A2,将被5除后余1的数字按给出顺序进行交错求和，即计算n1-n2+n3-n4...
x=x*q;
q=q*(-1);
b=b+x;
}
if(x%5==2) {//A3,被5除后余2的数字个数
c=c+1;
}
if(x%5==3) {//A4,被5除后余3的数字的平均数，精确到小数点后1位
d=d+x;
s=s+1;
}
if(x%5==4&&max<x) {
max=x;
}
}
if(d==0) {
s=1;
}
d=(double)d/(double)s;//当s=0的时候，语***错误
if(a==0){
System.out.print("N"+" ");
}else{
System.out.print(a+" ");
}
if(b==0){
System.out.print("N"+" ");
}else{
System.out.print(b+" ");
}
if(c==0){
System.out.print("N"+" ");
}else{
System.out.print(c+" ");
}
if(d==0){
System.out.print("N"+" ");
}else{
System.out.print(df.format(d)+" ");
}
if(max==0){
System.out.print("N");
}else{
System.out.print(max);
}
}
}
发表于 2018-04-13 17:08:35
回复(0)
import java.text.DecimalFormat;
import java.util.Scanner;
public class Main {
public static void main(String [] args){
Scanner s=new Scanner(System.in);
int n=s.nextInt();
int a1=0,a2=0,a3=0,a4=0,a5=0;
boolean b1=false,b2=false,b3=false,b4=false,b5=false;
int []a=new int[n];
int factor=1;
int k=0;
String c4=" ";
for(int i=0;i<n;i++)
{
a[i]=s.nextInt();
}
for(int i=0;i<n;i++)
{
switch(a[i]%5){
case 0:
if(a[i]%2==0) {
a1+=a[i];
b1=true;
}
break;
case 1:
a2+=a[i]*factor;
factor=-factor;
b2=true;
break;
case 2:
a3++;
b3=true;
break;
case 3:
a4+=a[i];
k++;
b4=true;
break;
case 4:
if(a5<a[i]) {
a5=a[i];
b5=true;
}
break;
}
}
if(b1)
System.out.print(a1+" ");
else
System.out.print("N ");
if(b2)
System.out.print(a2+" ");
else
System.out.print("N ");
if(b3)
System.out.print(a3+" ");
else
System.out.print("N ");
if(b4) {
c4 = new DecimalFormat("0.0").format(a4*1.0/k);
System.out.print(c4+" ");
}
else
System.out.print("N ");
if(b5)
System.out.print(a5);
else
System.out.print("N");
}
}
编辑于 2018-01-30 21:37:17
回复(0)
import java.util.Scanner;
public class Main {
private final static Scanner input = new Scanner(System.in);
public static void main(String[] args) {
int number = input.nextInt();
float[] result = new float[5];
int exchange = 1, count = 0;
for (int i = 0; i < number; i++) {
int temp = input.nextInt();
if (temp % 5 == 0) {
if (temp % 2 == 0) {
result[0] += temp;
}
}
if (temp % 5 == 1) {
result[1] += (temp * exchange);
exchange = -exchange;
}
if (temp % 5 == 2) {
++result[2];
}
if (temp % 5 == 3) {
result[3] += temp;
++count;
}
if (temp % 5 == 4) {
if (result[4] < temp) {
result[4] = temp;
}
}
}
for (int i = 0; i < result.length; i++) {
if (result[i] == 0) {
System.out.println((i==4)?"N":"N ");
} else if (i == 3) {
System.out.print(String.format("%.1f", result[i] / count) + " ");
} else if ((i == 4)) {
System.out.print((int) result[i]);
}else {
System.out.print((i==4)?(int) result[i]:(int) result[i] + " ");
}
}
}
}
发表于 2017-11-23 01:05:13
回复(0)
自己都有点嫌弃这个代码呢，不过可以AC~~~import java.util.Scanner;
public class Main {
public static void main(String[] args) {
@SuppressWarnings("resource")
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int A1 = 0;// A1也就是能被10整除的数字之和
int A2 = 0;// A2也就是将被5除后余1的数字按给出顺序进行交错求和
double A4 = 0;// 被5除后余3的数字的平均数，精确到小数点后1位
int A2TemporaryVariable = 0;
int A3TemporaryVariable = 0;
int A4TemporaryVariable = 0;
int maxNum = 0;
for (int i = 0; i < n; i++) {
// 开始一个一个输入，一个一个进行相应的计算
int num = in.nextInt();
if (num % 10 == 0) {
A1 += num;
} else if (num % 5 == 1) {
A2 += Math.pow(-1, A2TemporaryVariable) * num;
A2TemporaryVariable++;
} else if (num % 5 == 2) {
A3TemporaryVariable++;
} else if (num % 5 == 3) {
A4 += num;
A4TemporaryVariable++;
} else if (num % 5 == 4) {
if (num > maxNum) {
maxNum = num;
}
}
}
if (A1 == 0) {
System.out.print("N" + " ");
} else {
System.out.print(A1 + " ");
}
if (A2 == 0) {
System.out.print("N" + " ");
} else {
System.out.print(A2 + " ");
}
if (A3TemporaryVariable == 0) {
System.out.print("N" + " ");
} else {
System.out.print(A3TemporaryVariable + " ");
}
if (A4TemporaryVariable == 0) {
System.out.print("N" + " ");
} else {
System.out.print(String.format("%.1f" + " ", A4 / A4TemporaryVariable));
}
if (maxNum == 0) {
System.out.print("N");
} else {
System.out.print(maxNum);
}
}
}
编辑于 2017-09-18 16:21:41
回复(0)
| importjava.util.*; importjava.math.BigDecimal; importjava.text.DecimalFormat; publicclassMain{ publicstaticvoidmain(String[] args){ Scanner input=newScanner(System.in); intN=input.nextInt(); intk[]=newint[N]; intA[]=newint[5]; intsum[]=newint[5]; intt=0; for(inti=0;i<N;i++){ k[i]=input.nextInt(); } for(inti=0;i<N;i++){ if(k[i]%5==0&&k[i]%2==0){ sum[0]++; A[0]+=k[i]; }elseif(k[i]%5==1){ sum[1]++; A[1]=A[1]+(int)Math.pow(-1,t)*k[i]; t++; }elseif(k[i]%5==2){ sum[2]++; A[2]++; }elseif(k[i]%5==3){ sum[3]++; A[3]+=k[i]; }elseif(k[i]%5==4){ sum[4]++; A[4]=A[4]<k[i]?k[i]:A[4]; } } for(inti=0;i<4;i++){ if(sum[i]==0){ System.out.print("N "); } elseif(i!=3){ System.out.print(A[i]+" "); }else{ doublef=(double)A[i]/(double)sum[i]; DecimalFormat df=newDecimalFormat(); BigDecimal b = newBigDecimal(f); doublef1 = b.setScale(1,BigDecimal.ROUND_HALF_UP).doubleValue(); System.out.print(f1+" "); } } if(sum[4]==0){ System.out.print("N"); }else{ System.out.print(A[4]); } } } |
发表于 2017-09-15 01:35:31
回复(0)
一开始为了省事用了数组，实际上用基本类型代码会更少。
本题的测试用例没有考虑到偶数个被5除余1的数的边界情况，比如测试用例"2 5 5”。直接判断A2==0然后输出N属于投机取巧，出题的锅。
另外一开始没有注意题干说的是正整数，设计数组f存放标志其实可以不用的。
如有错误，欢迎指正。
import java.math.RoundingMode;
import java.text.DecimalFormat;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
int[] a = new int[5];
boolean[] f = new boolean[5];
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int j = 0;
int n2 = 0;
int n4 = 0;
while (j++ < n) {
int i = sc.nextInt();
int r = i % 5;
switch (r) {
case 0:
if (i % 2 == 0) {
f[0] = true;
a[0] += i;
}
break;
case 1:
f[1] = true;
if (++n2 % 2 == 0) {
a[1] -= i;
} else {
a[1] += i;
}
break;
case 2:
f[2] = true;
a[2]++;
break;
case 3:
f[3] = true;
a[3] += i;
n4++;
break;
case 4:
f[4] = true;
a[4] = Math.max(a[4], i);
break;
default:
break;
}
}
StringBuffer sb = new StringBuffer();
for (int i = 0; i < 5; i++) {
if (f[i] == false) {
sb.append("N");
} else {
if (i != 3) {
sb.append(a[i]);
} else {
DecimalFormat df = new DecimalFormat("0.0");
df.setRoundingMode(RoundingMode.HALF_UP);
sb.append(df.format(((double) a[3]) / n4));
}
}
sb.append(" ");
}
sb.deleteCharAt(sb.length()-1);
System.out.println(sb.toString());
}
}
编辑于 2017-08-23 16:21:27
回复(0)
//输出有些不太好，但是自己的第一想法是这个，所以就没更改，刚看了评论区，
//有System.out.print(a3==0?"N ":a3+" ");这种方法。书写更简单
import java.util.Scanner;
public class Mian{
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int num = sc.nextInt();//获取第一个值
int[] a = new int[num];
int[] tem = new int[num];
for(int i = 0; i < num;i++){
tem[i] = sc.nextInt();
a[i] = tem[i]%5;
}
int A1 = 0,A2 = 0,A22 = 0,A3 = 0,A44 = 0,A5 = 0;
double A4 = 0.0;
boolean a1 = false,a2 = false,a3 = false,a4 = false,a5 = false;
for(int i = 0; i < num;i++){
switch(a[i]){
case 0 : if(tem[i] % 2 == 0){ A1 += tem[i]; a1 = true;}break;
case 1 : { a2 = true;
if(A22%2 == 0)
A2 += tem[i];
else
A2 -= tem[i];
}A22++;break;
case 2 : a3 = true; A3++;break;
case 3 : a4 = true; A44++;A4 += tem[i];break;
case 4 : a5 = true; if(tem[i] > A5)A5 = tem[i];break;
}
}
if(a1)
System.out.print(A1+" ");
else
System.out.print("N ");
if(a2)
System.out.print(A2+" ");
else
System.out.print("N ");
if(a3)
System.out.print(A3+" ");
else
System.out.print("N ");
if(a4)
System.out.printf("%.1f ",A4/A44);
else
System.out.print("N ");
if(a5)
System.out.print(A5);
else
System.out.print("N");
}
}
发表于 2017-08-20 21:41:21
回复(0)
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int[] arr = new int[N];
int sum = 0;// 能被5整除的数字中所有偶数的和
int sum1 = 0;//将被5除后余1的数字按给出顺序进行交错求和
int sum2 = 0;//被5除后余2的数字的个数
int sum3 = 0;//被5除后余3的数字的和
int max = 0;//被5除后余4的数字中最大数字
int i = 0;
int j = 1;//对将被5除后余1的数字进行标记
int num = 0;//被5除后余3的数字的个数
while (i < N){
arr[i] = sc.nextInt();
int y = arr[i]%5;
switch (y){
case 0:
if (arr[i]%2 == 0){
sum += arr[i];
}
break;
case 1:
if(j % 2 == 1){
sum1 += arr[i];
} else {
sum1 -= arr[i];
}
j++;
break;
case 2:
sum2++;
break;
case 3:
sum3 += arr[i];
num++;
break;
case 4:
if(arr[i] > max){
max = arr[i];
}
break;
}
i++;
}
if(sum == 0){
System.out.print("N ");
}else {
System.out.print(sum+" ");
}
if(j == 1){
System.out.print("N ");
}else {
System.out.print(sum1+" ");
}
if(sum2 == 0){
System.out.print("N ");
}else {
System.out.print(sum2+" ");
}
if(num == 0){
System.out.print("N ");
}else {
System.out.print(String.format("%.1f",(double)sum3/num)+" ");
}
if(max == 0){
System.out.print("N");
}else{
System.out.print(max);
}
}
}
编辑于 2017-05-06 11:35:03
回复(0)
思路不难，关键是实现（Java），整理一下我遇到的问题：
1.数据的存储：为了节省空间，采用List类型
1.1定义：
List<数据类型> list=new Arraylist<数据类型>( ); or
List<数据类型> list=new Linkedlist<数据类型>；
1.2区别：
ArrayList
和Vector是采取数组体式格式存储数据，此数组元素数大于实际存储的数据以便增长和插入元素，都容许直接序号索引元素，然则插入数据要设计到数组元素移动等内存操纵，所以索引数据快插入数据慢，Vector因为应用了synchronized办法（线程安然）所以机能上比ArrayList要差，LinkedList应用双向链表实现存储，按序号索引数据须要进行向前或向后遍历，然则插入数据时只须要记录本项的前后项即可，所以插入数度较快！
1.3用法：
添加元素：list.add(x); or list.add(index,x);
删除元素：list.remove(x); or list.remove(index);
替换元素：list.set(index,x);//将指定元素替换为x
取出元素：list.get(index);
清空集合：list.clear();
2.保留指定位数的小数
double a4；//需要转化的数
BigDecimal bi=new BigDecimal(a4);
double
a=bi.setScale(1,BigDecimal.ROUND_HALF_UP).doubleValue();//四舍五入，保留一位小数
3.System.out.println()输出后自动换行，而System.out.print()输出后不自动换行
代码：
/*
*author:Weily
*time:2017.4.24
* */
import java.util.*;
import java.math.*;
@SuppressWarnings("unchecked")
public class test1{
public static void main(String[] srgs) {
int N;
int a1=0,a3=0,a2=0,a5=0;
double a4=0,a4out=0;
Scanner sc=new Scanner(System.in);
N=sc.nextInt();
int[] a=new int[N];
for(int i=0;i<N;i++)
a[i]=sc.nextInt();
List<Integer> A1=new ArrayList<Integer>();
List<Integer> A2=new ArrayList<Integer>();
List<Integer> A3=new ArrayList<Integer>();
List<Integer> A4=new ArrayList<Integer>();
List<Integer> A5=new ArrayList<Integer>();
try{
for(int i=0;i<N;i++){
if(a[i]%5==0) A1.add(a[i]);
else if(a[i]%5==1) A2.add(a[i]);
else if(a[i]%5==2) A3.add(a[i]);
else if(a[i]%5==3) A4.add(a[i]);
else A5.add(a[i]);
}
}catch(Exception e){}
try{
if(A1.isEmpty()) System.out.print("N"+" ");
else {
for(int i=0;i<A1.size();i++){
if((int)A1.get(i)%2==0)
a1=a1+(int)A1.get(i);
}
if(a1==0)System.out.print("N"+" ");
else
System.out.print(a1+" ");
}
if(A2.isEmpty()) System.out.print("N"+" ");
else {
for(int i=0;i<A2.size();i++)
a2=a2+(int)Math.pow(-1, i)*A2.get(i);
System.out.print(a2+" ");
}
if(A3.isEmpty()) System.out.print("N"+" ");
else {
a3=A3.size();
System.out.print(a3+" ");
}
if(A4.isEmpty()) System.out.print("N"+" ");
else{
double sum=0;
for(int i=0;i<A4.size();i++) sum+=(int)A4.get(i);
a4=sum/A4.size();
BigDecimal bi=new BigDecimal(a4);
a4out=bi.setScale(1,BigDecimal.ROUND_HALF_UP).doubleValue();
System.out.print(a4out+" ");
}
if(A5.isEmpty()) System.out.print("N");
else{
int max=0;
for(int i=0;i<A5.size();i++)
if((int)A5.get(i)>max) max=(int)A5.get(i);
a5=max;
System.out.print(a5);
}
}catch(Exception e){ }
}
代码：
/*
*author:Weily
*time:2017.4.24
* */
import java.util.*;
import java.math.*;
@SuppressWarnings("unchecked")
public class test1{
public static void main(String[] srgs) {
int N;
int a1=0,a3=0,a2=0,a5=0;
double a4=0,a4out=0;
Scanner sc=new Scanner(System.in);
N=sc.nextInt();
int[] a=new int[N];
for(int i=0;i<N;i++)
a[i]=sc.nextInt();
List<Integer> A1=new ArrayList<Integer>();
List<Integer> A2=new ArrayList<Integer>();
List<Integer> A3=new ArrayList<Integer>();
List<Integer> A4=new ArrayList<Integer>();
List<Integer> A5=new ArrayList<Integer>();
try{
for(int i=0;i<N;i++){
if(a[i]%5==0) A1.add(a[i]);
else if(a[i]%5==1) A2.add(a[i]);
else if(a[i]%5==2) A3.add(a[i]);
else if(a[i]%5==3) A4.add(a[i]);
else A5.add(a[i]);
}
}catch(Exception e){}
try{
if(A1.isEmpty()) System.out.print("N"+" ");
else {
for(int i=0;i<A1.size();i++){
if((int)A1.get(i)%2==0)
a1=a1+(int)A1.get(i);
}
if(a1==0)System.out.print("N"+" ");
else
System.out.print(a1+" ");
}
if(A2.isEmpty()) System.out.print("N"+" ");
else {
for(int i=0;i<A2.size();i++)
a2=a2+(int)Math.pow(-1, i)*A2.get(i);
System.out.print(a2+" ");
}
if(A3.isEmpty()) System.out.print("N"+" ");
else {
a3=A3.size();
System.out.print(a3+" ");
}
if(A4.isEmpty()) System.out.print("N"+" ");
else{
double sum=0;
for(int i=0;i<A4.size();i++) sum+=(int)A4.get(i);
a4=sum/A4.size();
BigDecimal bi=new BigDecimal(a4);
a4out=bi.setScale(1,BigDecimal.ROUND_HALF_UP).doubleValue();
System.out.print(a4out+" ");
}
if(A5.isEmpty()) System.out.print("N");
else{
int max=0;
for(int i=0;i<A5.size();i++)
if((int)A5.get(i)>max) max=(int)A5.get(i);
a5=max;
System.out.print(a5);
}
}catch(Exception e){ }
}
编辑于 2017-04-25 00:05:34
回复(0)
问题信息
通过挑战的用户
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2023-03-13 16:55:58
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