Code
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1306 Accepted Submission(s): 540
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;
}
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).
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.
Print an integer,denoting what the function returns.
Sample Input
5
1 3 4 2 4
Sample Output
64
Hint
gcd(x,y) means the greatest common divisor of x and y.题意：给定序列1≤i,j≤n，求gcd(a[i],a[j])∗(gcd(a[i],a[j])−1)之和。
思路：倍数莫比乌斯反演。
代码：
1 #include<bits/stdc++.h> 2 #define ll long long 3 using namespace std; 4 const int N = 1e5 + 5; 5 const int mod=10007; 6 int t; 7 //线性筛法求莫比乌斯函数 8 bool vis[N + 10]; 9 int pri[N + 10]; 10 int mu[N + 10]; 11 int sum[N]; 12 13 void mus() { 14 memset(vis, 0, sizeof(vis)); 15 mu[1] = 1; 16 int tot = 0; 17 for (int i = 2; i < N; i++) { 18 if (!vis[i]) { 19 pri[tot++] = i; 20 mu[i] = -1; 21 } 22 for (int j = 0; j < tot && i * pri[j] < N; j++) { 23 vis[i * pri[j]] = 1; 24 if (i % pri[j] == 0) { 25 mu[i * pri[j]] = 0; 26 break; 27 } 28 else mu[i * pri[j]] = -mu[i]; 29 } 30 } 31 sum[1]=1; 32 for(int i=2;i<N;i++) sum[i]=sum[i-1]+mu[i]; 33 } 34 int n,m,k; 35 36 int a[N]; 37 int b[N]; 38 int F[N]; 39 int main() { 40 mus(); 41 while(scanf("%d",&n)==1){ 42 int ma=0; 43 memset(F,0, sizeof(F)); 44 memset(b,0, sizeof(b)); 45 for(int i=0;i<n;i++){ 46 scanf("%d",&a[i]); 47 b[a[i]]++;//b[a[i]]的个数 48 ma=max(ma,a[i]); 49 } 50 for(int i=1;i<=ma;i++) 51 for(int j=i;j<=ma;j+=i) F[i]+=b[j];//在范围内i的倍数的个数 52 ll ans=0; 53 for(int i=1;i<=ma;i++){ 54 ll res=0; 55 for(int j=i;j<=ma;j+=i){ 56 if(!F[j]) continue; 57 res+=1ll*mu[j/i]*F[j]*F[j]%mod;//公约数均为i 58 } 59 ans=(ans+res*i%mod*(i-1))%mod;//同时乘上i*i-1 60 } 61 ans%=mod; 62 ans+=mod; 63 ans%=mod; 64 printf("%lld ",ans); 65 } 66 return 0; 67 }