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Code
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 77 Accepted Submission(s): 27
Problem Description WLD likes playing with codes.One day he is writing a function.Howerver,his computer breaks down because the function is too powerful.He is very sad.Can you help him?
The function:
int calc
{
int res=0;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
{
res+=gcd(a[i],a[j])*(gcd(a[i],a[j])-1);
res%=10007;
}
return res;
}
Input There are Multiple Cases.(At MOST
10
)
For each case:
The first line contains an integer
N(1≤N≤10000)
.
The next line contains
N
integers
a1,a2,...,aN(1≤ai≤10000)
.
Output For each case:
Print an integer,denoting what the function returns.
Sample Input
5
1 3 4 2 4
Sample Output
64
Hintgcd(x,y) means the greatest common divisor of x and y.
Source BestCoder Round #39 ($)
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给你一个a数组，让你计算那段代码的结果，用容斥原理就ok，莫比乌斯反演也行，不过不会莫比乌斯反演（也是容斥原理）orz。。。改天学习下。
代码如下：
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#include
#include
#include
#include
#include
#include