C++练习一（poj练习）

1. 练习一poj.org的1000题

题目：

• Description
  Calculate a+b
• Input
  Two integer a,b (0<=a,b<=10)
• Output
  Output a+b
• Sample Input
  1 2
• Sample Output
  3

代码：

#include<iostream>
using namespace std;
int main()
{
	int a,b;
	cout<<"a(a>=0):  ";
	cin>>a;
	if(a<0){
		cout<<"请重新输入： ";
		cout<<"a(a>=0):";
	}
	cout<<"b(b<=10):  ";
	cin>>b;
		if(b>10){
		cout<<"请重新输入： ";
		cout<<"b(b<=10):  ";
	}
	
	cout<<"a+b="<<a+b;
	return 0;
 }

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2. 练习二

题目：

• Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
• Example:
  Input: 38
  Output: 2
  Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2.

  Since 2 has only one digit, return it.

代码：
第一种方法：

//使用递归方法
#include<cmath>
using namespace std;

class Solution {
public:
    int addDigits(int num) {
    int index=0;
    while(num!=0)
    {
    	index+=num%10;
    	num=num/10;
	}
	num=index;
	if(num>=10)
    return addDigits(num);
    else
	return num;
	
}
};
int main()
{
	Solution s;
	cout<<s.addDigits(120)<<endl;
}

第二种方法：

//使用迭代法
#include<iostream>
#include<cmath>
using namespace std;
class Solution {
public:
    int addDigits(int num) {
    int index=0,flag=0;
    while(num>=10)
    {   flag=num;
    	while(flag!=0)
    	{
    	index+=flag%10;
    	flag=flag/10;
        }
        num=index;
        index=0;
	}
	return num;
	
}
};
int main()
{
	Solution s;
	cout<<s.addDigits(120)<<endl;
}

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3. 练习三poj.org的1004题

题目：

• Description
  Larry graduated this year and finally has a job. He’s making a lot of money, but somehow never seems to have enough. Larry has decided that he needs to grab hold of his financial portfolio and solve his financing problems. The first step is to figure out what’s been going on with his money. Larry has his bank account statements and wants to see how much money he has. Help Larry by writing a program to take his closing balance from each of the past twelve months and calculate his average account balance.
• Input
  The input will be twelve lines. Each line will contain the closing balance of his bank account for a particular month. Each number will be positive and displayed to the penny. No dollar sign will be included.
• Output
  The output will be a single number, the average (mean) of the closing balances for the twelve months. It will be rounded to the nearest penny, preceded immediately by a dollar sign, and followed by the end-of-line. There will be no other spaces or characters in the output.
• Sample Input

100.00
489.12
12454.12
1234.10
823.05
109.20
5.27
1542.25
839.18
83.99
1295.01
1.75

• Sample Output
  $1581.42

代码：

#include<iostream>>
using namespace std;
int main()
{
	float month,sum=0;
	for(int i=0;i<12;i++)
	{
		cin>>month;
		sum+=month;
	}
	cout<<"$"<<sum/12;
	return 0;	
}

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4. 练习四poj.org的3980题

题目：

• Description
  编写一个C函数mod(int n, int m),实现取模运算%
• Input
  输入包含多行数据
  每行数据是两个整数a, b (1 <= a, b <= 32767)
  数据以EOF结束
• Output
  于输入的每一行输出a%b
• Sample Input
  5 3
  100 2
• Sample Output
  2
  0

代码：

#include<iostream>
#include<string>
using namespace std;
int mod(int n,int m)
{
   return n-(n/m)*m;
}
int main()
{
int a,b;
string ch;
    while(cin>>a>>b)
    { 
	cout<<mod(a,b)<<endl;
	if((ch=cin.get())=="EOF")	
		break;
    }

}

注意：
取模与取余的区别：

• 取余运算在计算商值向0方向舍弃小数位
• 取模运算在计算商值向负无穷方向舍弃小数
• 取余，遵循尽可能让商大的原则
• 取模，遵循尽可能让商小的原则

从上面的区别可以总结出，取余（rem）和取模（mod）在被除数、除数同号时，结果是等同的，异号时会有区别，所以要特别注意异号的情况,经过测试，在C/C++, C#, JAVA, PHP这几门主流语言中，’%’运算符都是做取余运算，而在python中的’%’是做取模运算。在C/C++, C#, JAVA, PHP中a对b取模的公式为a-(a/b)*b。

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