一道JAVA题:试读入一个16位的长整型,利用switch语句计算0—9这十个数字中,每个字出现的次数? 在Java语言中怎样给一个长整型数保留两位?

\u8bd5\u8bfb\u5165\u4e00\u4e2a16\u4f4d\u7684\u957f\u6574\u578b\uff0c\u5229\u7528switch\u8bed\u53e5\u8ba1\u7b970\u20149\u8fd9\u5341\u4e2a\u6570\u5b57\u4e2d\uff0c\u6bcf\u4e2a\u5b57\u51fa\u73b0\u7684\u6b21\u6570

#include
main(){
int a[10],i;
long x;
scanf("%ld",&x);
for (i=0;i<=9;i++) a[i]=0;
while(x)
{
a[x%10]++;
x/=10;
}
for (i=0;i<=9;i++) printf("%d:%d\n",i,a[i]);
}

\u8c03\u8bd5\u901a\u8fc7\uff0c\u8fd0\u884c\u6b63\u786e

double result;
long i= 1000000;
result=(i*100/1024)/(double)100;
\u957f\u6574\u578b\u6570\u6ca1\u6709\u5c0f\u6570\u4f4d\uff0c\u8981\u4fdd\u7559\u5c0f\u6570\u53ea\u80fd\u628a\u7ed3\u679c\u4fdd\u5b58\u5728double\uff08/float\uff09\u578b\u6570\u4e2d\u2026\u2026

思路:先把16位的长整型转换成字符串,再将字符串拆成字符数组,对字符数组进行switch,利用一个int数组来储存每个数字出现的次数。

具体实现:
文件名:CountNum.java

import java.util.Scanner;

public class CountNum {
public static void main(String[] args){
System.out.println("请输入一个16位数字");
Scanner sc = new Scanner(System.in);
long l = sc.nextLong();
String s = l+"";
char[] ca = new char[16];
s.getChars(0, s.length(), ca, 0);
int[] ia = new int[10];
for(int i=0;i<ca.length;i++){
switch(ca[i]){
case '0': ia[0]++;break;
case '1': ia[1]++;break;
case '2': ia[2]++;break;
case '3': ia[3]++;break;
case '4': ia[4]++;break;
case '5': ia[5]++;break;
case '6': ia[6]++;break;
case '7': ia[7]++;break;
case '8': ia[8]++;break;
case '9': ia[9]++;break;
}
}
System.out.println("统计结果是:");
for(int i=0;i<ia.length;i++){
System.out.println("数字"+i+"出现了"+ia[i]+"次。");
}
}
}

foreach(int i in 16位长整型){
switch(i){
case 0:
。。。。。
}
}
大概思路就是这样的了。具体怎么实现自己想想吧。因为涉及变量太多,没时间

public static void main(String[] args) {
Scanner sca = new Scanner(System.in);
long l = sca.nextLong();
int[] ints = new int[10];
String str = l + "";
for (char c : str.toCharArray()) {
ints[(int) c - 48]++;
}
for (int i = 0; i < 10; i++) {
System.out.println(i + "出现的次数为:" + ints[i]);
}
}

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