hdu4473(排列组合)

Exam

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 497    Accepted Submission(s): 203

Problem Description

Rikka is a high school girl suffering seriously from Chūnibyō (the age of fourteen would either act like a know-it-all adult, or thinks they have special powers no one else has. You might google it for detailed explanation) who, unfortunately, performs badly at math courses. After scoring so poorly on her maths test, she is faced with the situation that her club would be disband if her scores keeps low.
Believe it or not, in the next exam she faces a hard problem described as follows.
Let’s denote f(x) number of ordered pairs satisfying (a * b)|x (that is, x mod (a * b) = 0) where a and b are positive integers. Given a positive integer n, Rikka is required to solve for f(1) + f(2) + . . . + f(n).
According to story development we know that Rikka scores slightly higher than average, meaning she must have solved this problem. So, how does she manage to do so?

 

Input

There are several test cases.
For each test case, there is a single line containing only one integer n (1 ≤ n ≤ 10¹¹).
Input is terminated by EOF.

 

Output

For each test case, output one line “Case X: Y” where X is the test case number (starting from 1) and Y is the desired answer.

 

Sample Input

13610152128

 

Sample Output

Case 1: 1Case 2: 7Case 3: 25Case 4: 53Case 5: 95Case 6: 161Case 7: 246

 

Source

2012 Asia Chengdu Regional Contest

 

对于f[x],xmod(a*b)=0,即存在c,使a*b*c=x,f[x]即为这样的a,b的组合个数;题目要求的是f[1]+f[2]+f[3]+…+f[n-1]+f[n]的和，且n很大。题目数据之大，打消了直接枚举的念头，必须用算法降低时间复杂度。

f[1]+f[2]+f[3]+…+f[n-1]+f[n]的和即可这样理解：a*b*c<=n，求这样的a,b,c的组合个数。（若将a、b、c按递增顺序排列）

    A、a=b=c,

    B、a=b<c

    C、a<b=c

    D、a<b<c

由于题目abc可以任意排列，故最终答案为A+（B+C）*3+D*6,剩下的任务就是求ABCD了。

 

#include<iostream>#include<cmath>#include<cstdio>using namespace std;int main(){__int64 n,ans,i,j,a,b,tag=1;while(~scanf("%I64d",&n)){ans=0;a=pow((double)n,1.0/3.0);while(a*a*a<n)a++;if(a*a*a>n)a--;ans+=a;for(i=1;i<=a;i++)///枚举a的可能取值{__int64 tmp=n/i;b=sqrt(tmp);while(b*b<tmp)b++;if(b*b>tmp)b--;ans+=(b-i+tmp/i-i)*3;for(j=i+1;j<=b;j++)ans+=(tmp/j-j)*6;}printf("Case %I64d: %I64d\n",tag++,ans);}return 0;}