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B. Easy Number Challenge

2015-07-20 17:37:21 来源: 作者: 【大中小】浏览:3次

B. Easy Number Challenge

time limit per test 2 seconds
memory limit per test 256 megabytes

题目链接:传送门

Let's denote d(n) as the number of divisors of a positive integern. You are given three integers a, b and c. Your task is to calculate the following sum:

$\$

Find the sum modulo 1073741824(2³⁰).

Input

The first line contains three space-separated integersa, b andc (1?≤?a,?b,?c?≤?100).

Output

Print a single integer ― the required sum modulo 1073741824 (2³⁰).

Sample test(s) Input

2 2 2

Output

Input

5 6 7

Output

Note

For the first example.

d(1?1?1)?=? d(1)?=?1; d(1?1?2)?=? d(2)?=?2; d(1?2?1)?=? d(2)?=?2; d(1?2?2)?=? d(4)?=?3; d(2?1?1)?=? d(2)?=?2; d(2?1?2)?=? d(4)?=?3; d(2?2?1)?=? d(4)?=?3; d(2?2?2)?=? d(8)?=?4.

So the result is 1?+?2?+?2?+?3?+?2?+?3?+?3?+?4?=?20.

意解:暴力,先把暴力枚举一边,记录每个数出现的个数,然后Nsqrt(N)找因子;

AC代码:

#include 
  
   
#include 
   
     #include 
    
      #include 
     
       #define For(i,a,n) for(int i = a; i <= n; i++) using namespace std; const int M = 1e6; const int INF = 1LL << 30; int ma[M + 10]; int main() { int a,b,c; int ans = 0; scanf("%d %d %d",&a,&b,&c); For(i,1,a) For(j,1,b) For(k,1,c) ma[i * j * k]++; for(int i = 1; i <= M; i++) { if(ma[i]) for(int p = 1; p * p <= i; p++) { if(i % p == 0) { ans += ma[i]; if(p * p != i) ans += ma[i]; } } } cout<


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