Description
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mmoaay小侄子今年上初中,老師出了一道求約數個數的題目,比如8的約數有1,2,4,8共4個。
當數比較小的時候可以人工算,當n較大時就難了。
mmoaay嫌麻煩,現在讓你編個程序來算。
Input
一行一個整數。最后以0結束。
Output
分別求出這些整數的約數個數,最后的0不用處理。
Sample Input
8
100
0
Sample Output
4
9
一個知識點:任何數都可以寫成2的p1次方乘以3的p2次方乘以5的p3次方……也就是素數的乘積 ?約數的個數是(p1+1)*(p2+1)……
AC代碼
#include <iostream>
#include<stdio.h>
using namespace std;
#include <string.h>
int main()
{int n,i;while(1){scanf("%d",&n);if(n==0)break;int d=1;for(i=2;n!=1;i++){int c=1;while(n%i==0){c++;n/=i;}d*=c;}printf("%d\n",d);}return 0;
}
The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the question of primality. A prime number is a number that is has no proper factors (it is only evenly divisible by 1 and itself). The first prime numbers are 2,3,5,7 but they quickly become less frequent. One of the interesting questions is how dense they are in various ranges. Adjacent primes are two numbers that are both primes, but there are no other prime numbers between the adjacent primes. For example, 2,3 are the only adjacent primes that are also adjacent numbers.?
Your program is given 2 numbers: L and U (1<=L< U<=2,147,483,647), and you are to find the two adjacent primes C1 and C2 (L<=C1< C2<=U) that are closest (i.e. C2-C1 is the minimum). If there are other pairs that are the same distance apart, use the first pair. You are also to find the two adjacent primes D1 and D2 (L<=D1< D2<=U) where D1 and D2 are as distant from each other as possible (again choosing the first pair if there is a tie).
Input
Each line of input will contain two positive integers, L and U, with L < U. The difference between L and U will not exceed 1,000,000.
Output
For each L and U, the output will either be the statement that there are no adjacent primes (because there are less than two primes between the two given numbers) or a line giving the two pairs of adjacent primes.
Sample Input
2 17
14 17
Sample Output
2,3 are closest, 7,11 are most distant.
There are no adjacent primes.
題意:判斷相鄰的素數差最小與差最大