HDOJ 4473 Exam 暴力枚举

所求的既 a*b*c<=n 的对数.....

暴力枚举a,b根据c计算结果

Exam

Time Limit: 20000/10000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 1316    Accepted Submission(s): 559


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

1
3
6
10
15
21
28

 

Sample Output

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

 

Source

2012 Asia Chengdu Regional Contest

 

/* ***********************************************
Author :CKboss
Created Time :2015年09月02日 星期三 12时42分31秒
File Name :HDOJ4473.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef unsigned long long int LL;

LL n;

int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout);

int cas=1;
while(scanf("%lld",&n)!=EOF)
{
LL ans=0;
for(LL a=1;a*a<=n;a++)
{
for(LL b=a;b*b*a<=n;b++)
{
LL c=n/(a*b);
LL ret=0LL;
if(a==b) ret+=1LL;
else ret+=3LL;
if(c>b) { if(a==b) ret+=3LL*(c-b); else ret+=6LL*(c-b); }

ans+=ret;
}
}
printf("Case %d: %lld\n",cas++,ans);
}

return 0;
}