**最大公約數和最小公倍數
Write a program which computes the greatest common divisor (GCD) and the least common multiple (LCM) of given a and b.
Input
Input consists of several data sets. Each data set contains a and b separated by a single space in a line. The input terminates with EOF.
Constraints

0 < a, b ≤ 2,000,000,000
LCM(a, b) ≤ 2,000,000,000
The number of data sets ≤ 50

Output
For each data set, print GCD and LCM separated by a single space in a line.
Sample Input
8 6
50000000 30000000

Output for the Sample Input
2 24
10000000 150000000**

#include<stdio.h>
#include<math.h>
int main()
{
	int  a, b, x, y, t;
	int GCD, LCM;
	while (scanf("%d%d", &a, &b) != EOF)
	{
		x = a;
		y = b;
		if (a < b)
		{
			t = a;
			a = b;
			b = t;
		}
		while (b != 0)
		{
			t = b;
			b = a % b;
			a = t;
		}
		GCD = a;
		LCM = x / a * y;
		printf("%d %d\n", GCD, LCM);
	}
	return 0;
}
 

**素數
Write a program which reads an integer n and prints the number of prime numbers which are less than or equal to n. A prime number is a natural number which has exactly two distinct natural number divisors: 1 and itself. For example, the first four prime numbers are: 2, 3, 5 and 7.
Input
Input consists of several datasets. Each dataset has an integer n (1 ≤ n ≤ 999,999) in a line.
The number of datasets is less than or equal to 30.
Output
For each dataset, prints the number of prime numbers.
Sample Input
10
3
11

Output for the Sample Input
4
2
5
**

#include<stdio.h>
#include<string.h>
#include<iostream>
const int maxn = 1000005;
const int _maxn = 78499 + 5; 
int prime[_maxn];
bool vis[maxn];

int main()
{
	int n;
	while(scanf("%d",&n)!=EOF)
	{
		int count=0;
		memset(vis,0,sizeof(vis));
		for(int i=2;i<=n;i++)
		{
			if (!vis[i])  *vis[i]!=0時執行，也就是他是素數的時候執行*
				prime[count++] = i;
			for (int j = i+i; j<=n; j=j+i)
			{

				vis[j] = 1;*將不是素數的標記為1*
			}

		}
		printf("%d\n",count);
	}
	return 0;
}

**超級樓梯
有一樓梯共M級，剛開始時你在第一級，若每次只能跨上一級或二級，要走上第M級，共有多少種走法？

Input
輸入資料首先包含一個整數N，表示測試例項的個數，然後是N行資料，每行包含一個整數M（1<=M<=40）,表示樓梯的級數。

Output
對於每個測試例項，請輸出不同走法的數量
Sample Input
2
2
3

Sample Output
1
2**

#include<stdio.h>
int main()
{
	int n, m;
	int f[41];
	scanf("%d",&n);
         f[1] = 0;
	f[2] = 1;
	f[3] = 2;
		for(m=4;m<41;m++)
		{
			f[m] = f[m - 1] + f[m - 2];  *走一步或走兩步*
        }
		while(n--)
		{
			scanf("%d",&m);
			printf("%d\n",f[m]);
	     }
	return 0;
}

**計算球體積
根據輸入的半徑值，計算球的體積。

Input
輸入資料有多組，每組佔一行，每行包括一個實數，表示球的半徑。

Output
輸出對應的球的體積，對於每組輸入資料，輸出一行，計算結果保留三位小數。

Sample Input
1
1.5

Sample Output
4.189
14.137

Hint
#define PI 3.1415927

**

#include<stdio.h>
#include<math.h>
#define PI 3.1415927


int main()
{
	double r, v;
	while (scanf("%lf", &r) != EOF)
	{
		v = 4 * PI*r*r*r / 3;
		printf("%.3lf\n", v);
	}
	return 0;
}

**求實數的絕對值。

Input
輸入資料有多組，每組佔一行，每行包含一個實數。

Output
對於每組輸入資料，輸出它的絕對值，要求每組資料輸出一行，結果保留兩位小數。

Sample Input
123
-234.00

Sample Output
123.00
234.00**

#include<stdio.h>
int main()
{
	double a;
	while (scanf("%lf",&a)!=EOF)
	{
		if (a>=0)
		{
			printf("%.2lf\n", a);
		}
		else
		{
			printf("%.2lf\n", -a);
		}
		
	}
	return 0;
}