C++编程练习题大全（带答案）

}

for(int j=0;j

9. 编写函数，将给定的字符串的大写字母转变成小写字母； 其它字符不变。主函数输入一个字符串，调用该函数对其进行转换，分别输出原串和转换之后的串。 package moreEasy;

import java.util.Scanner;

public class The9th { static String str=\

public static void main(String[] args) { String str=\

Scanner cin=new Scanner(System.in); str=cin.nextLine(); zhuanHuan(str); }

public static void zhuanHuan(String str1){ str=str1;

str=str.toLowerCase(); System.out.println(str); } }

10. 编写一个函数，将字符数组a中的全部字符复制到字符数组b中. 不要使用strcpy函数. 主函数输入任意一个字符串，调该函数，复制出另一个字符串。将两个串输出。 package moreEasy;

import java.util.Scanner; public class The10th {

public static void main(String[] args) { Scanner cin=new Scanner(System.in); String str=cin.nextLine();

char[] aa=new char[str.length()]; char[] bb=new char[str.length()]; for(int i=0;i

copy(aa,bb);

for(int i=0;i

409--17

System.out.println(\

for(int i=0;i

public static void copy(char[] a,char[] b){ for(int i=0;i

11. 判断一个N*N的矩阵是否为对称矩阵. 将原矩阵输出，判断结果输出. package moreEasy;

import java.util.Scanner; public class The11th {

public static void main(String[] args) { Scanner cin=new Scanner(System.in); int n=cin.nextInt(); int count=0;

int[][] duiCheng=new int[n][n]; for(int i=0;i

for(int j=0;j

for(int i=0;i

for(int j=0;j

System.out.print(duiCheng[i][j]+\ else

System.out.print(duiCheng[i][j]+\ }

for(int i=0;i

for(int j=0;j

if(count==n*n)

System.out.println(\对称矩阵\ else

System.out.println(\非对称矩阵\ }

409--18

}

12. 给出年、月、日, 计算该日是该年的第几天？ package moreEasy;

import java.util.Scanner; public class The12th {

public static void main(String[] args) { int n=0;

Scanner cin=new Scanner(System.in); int y=cin.nextInt();

System.out.println(\年\ int m=cin.nextInt();

System.out.println(\月\ int d=cin.nextInt();

System.out.println(\日\

if(y%4==0&&y0!=0||y@0==0){ switch(m){ case 1: n=d; break; case 2: n=d+31; break; case 3:

n=d+31+29; break; case 4:

n=d+31+29+31; break; case 5:

n=d+31+29+31+30; break; case 6:

n=d+31+29+31+30+31; break; case 7:

n=d+31+29+31+30+31+30; break; case 8:

n=d+31+29+31+30+31+30+31; break; case 9:

n=d+31+29+31+30+31+30+31+31; break; case 10:

n=d+31+29+31+30+31+30+31+31+30; break;

409--19

case 11:

n=d+31+29+31+30+31+30+31+31+30+31; break; case 12:

n=d+31+29+31+30+31+30+31+31+30+31+30; break; } switch(m){ case 1: n=d; break; case 2: n=d+31; break; case 3:

n=d+31+28; break; case 4:

n=d+31+28+31; break; case 5:

n=d+31+28+31+30; break; case 6:

n=d+31+28+31+30+31; break; case 7:

n=d+31+28+31+30+31+30; break; case 8:

n=d+31+28+31+30+31+30+31; break; case 9:

n=d+31+28+31+30+31+30+31+31; break; case 10:

n=d+31+28+31+30+31+30+31+31+30; break; case 11:

n=d+31+28+31+30+31+30+31+31+30+31; break; case 12:

n=d+31+28+31+30+31+30+31+31+30+31+30; break;

409--20

} else{