算數基本定理：

基本形式：（Pn表示質數）

導出結論（ACM中會用到的）：

1.????（正因數個數）

2.????（正因數之和）

例題：

1341 - Aladdin and the Flying Carpet

???

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Time Limit:?3 second(s)

Memory Limit:?32 MB

It's said that Aladdin had to solve seven mysteries before getting the Magical Lamp which summons a powerful Genie. Here we are concerned about the first mystery.

Aladdin was about to enter to a magical cave, led by the evil sorcerer who disguised himself as Aladdin's uncle, found a strange magical flying carpet at the entrance. There were some strange creatures guarding the entrance of the cave. Aladdin could run, but he knew that there was a high chance of getting caught. So, he decided to use the magical flying carpet. The carpet was rectangular shaped, but not square shaped. Aladdin took the carpet and with the help of it he passed the entrance.

Now you are given the area of the carpet and the length of the minimum possible side of the carpet, your task is to find how many types of carpets are possible. For example, the area of the carpet 12, and the minimum possible side of the carpet is 2, then there can be two types of carpets and their sides are: {2, 6} and {3, 4}.

Input

Input starts with an integer?T (≤ 4000), denoting the number of test cases.

Each case starts with a line containing two integers:?a?b?(1 ≤ b ≤ a ≤ 10¹²)?where?a?denotes the area of the carpet and?b?denotes the minimum possible side of the carpet.

Output

For each case, print the case number and the number of possible carpets.

Sample Input	Output for Sample Input
2 10 2 12 2	Case 1: 1 Case 2: 2

題意：給出a,b找出[b,a]中乘積為a的因子對個數（例如a=10,b=2,則只存在因子對{2,5}符合條件）

????????? 利用算數基本定理1求得[1,a]中因子對（求出因子數除以2），在暴力求解[1,b]中因子對，相減即可。

#include <cstdio> #include <iostream> #include <cmath> #include <cstring> #define MAX 1000047 typedef long long ll; using namespace std; ll a,b,p[MAX],prime[MAX],m,ans,t,cnt=0;void init() //標記素數并保存到prime數組中 {m=0;memset(p,0,sizeof(p));for (ll i=2;i<=MAX;i++){if (!p[i]){prime[m++]=i; //保存到prime數組 for (ll j=i+i;j<=MAX;j+=i)p[j]=1;}} }int main() {init();scanf("%d",&t);while(t--){scanf("%lld%lld",&a,&b);if (b>=sqrt(a)) //類似剪枝（否則會超時） {printf("Case %d: 0\n",++cnt);continue;}ll i=0,suma=1,sumb=0,x=a;while(prime[i]<a&&i<m) //為{if (a%prime[i]==0) //是因子{ll an=0;while(a%prime[i]==0){a/=prime[i];an++;} //求出指數 suma*=(an+1);}i++;}if (a>1) suma*=2;for (ll i=1;i<=sqrt(b);i++)if (x%i==0) sumb++; //找出[1,b)中的因子 ans=suma/2-sumb; //求出的是因子個數，除以2就是因子對個數 printf("Case %d: %lld\n",++cnt,ans);}return 0; }

核心代碼：

1.求素數（詳見https://blog.csdn.net/Radium_1209/article/details/80232925）

2.求各個指數

while(a%prime[i]==0) {a/=prime[i];an++; //an即為所求指數 }

3.帶入公式

suma*=(an+1);

轉載于:https://www.cnblogs.com/Radium1209/p/10415371.html

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