遵义市网站制作,电子商务的网站建设,竣工验收报告查询网,广西建设工程造价管理协会网站题目大意#xff1a;给两个数a#xff0c;b#xff0c;求满足c*da且cb且db的c,d二元组对数#xff0c;(c,d)和(d,c)属于同一种情况 题目分析#xff1a;根据唯一分解定理先将a唯一分解#xff0c;则a的所有正约数的个数为ans (1 a1) * (1 a2) *...(1 an) 因为…
题目大意给两个数ab求满足c*da且cb且db的c,d二元组对数(c,d)和(d,c)属于同一种情况 题目分析根据唯一分解定理先将a唯一分解则a的所有正约数的个数为ans (1 a1) * (1 a2) *...(1 an) 因为题目说了不会存在cd的情况因此ans要除2去掉重复情况。[care] 然后枚举小于b的a的约数拿ans减掉就可以了
Its 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 Aladdins 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
2
10 2
12 2
Sample Output
Case 1: 1
Case 2: 2
算数基本原理:任何一个大于1的自然数,都可以唯一分解成有限个质数的乘积 Np1^a1*p2^a2.....pn^an,这里p1p2...pn均为质数其诸指数是正整数。 定理应用1一个大于1的正整数N如果它的标准分解式为Np1^a1*p2^a2.....pn^an 那么它的正因数个数为f(n)(1a1)(1a2).....(1an)。
AC代码分步详解
#includeiostream
#includestring.h
#includemath.h
typedef long long ll;
#includestdio.h
using namespace std;
#define M 1000010
int dp[M];
int book[M];
int t,k;
ll m,n;
void dfs()
{k0;memset(dp,0,sizeof(dp));memset(book,0,sizeof(book));for(int i2; iM; i)/**因为任何一个整数都可以由一个素数经过乘法运算得到故可通过素数打表的方式求得某数的唯一分解得到幂次最大*/if(!book[i]){dp[k]i;/*记录素数*/for(int j2*i; jM; ji)book[j]1;}
}
int main()
{cint;int tt1;dfs();/*最好放在外面*/while(t--){cinmn;ll ans1;/**定义为 long long 型否则导致错误答案*/ll mmm;if(nsqrt(m))ans0;else{for(int i0; ik2*dp[i]m ; i) /**care:remember停止条件为ik2*dp[i]m缺一不可*/{if(m%dp[i]0){int a0;while(m%dp[i]0){m/dp[i];/**唯一分解定理直接对m的值进行操作得到xa1^b1*a2^b2.....an^bn*/a;}ansans*(a1);/**a的所有正约数的个数为ans (1 a1) * (1 a2) *...(1 an)*/}if(m1)break;}if(m1)/**for循环先判断导致停止故对最后出现的一个素数的情况判断*2*/ans*2;ans/2;///因为题目说了不会存在cd的情况因此ans要除2去掉重复情况。int b0;for(ll i1; in; i)/**i1,我的理解是前面由唯一分解定理求因数对个数时出现了1*m的情况当所有素数都为1时该情况出现此时减去*/if(mm%i0)/**care m的值在上面操作中发生改变故需要开变量存储m值*/b;ans-b;/**题目要求不出现小于n的因数对找出减去即可*/}printf(Case %d: %lld\n,tt,ans);}return 0;
}
