2015 Multi-University Training Contest 1 Hdu5288 OO’s Sequence

OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 449    Accepted Submission(s): 158


Problem Description

OO has got a array A of size n ,defined a function f(l,r) represent the number of i (l<=i<=r) , that there's no j(l<=j<=r,j<>i) satisfy ai mod aj=0,now OO want to know

∑i=1n∑j=inf(i,j) mod （109+7）.

 

Input

There are multiple test cases. Please process till EOF.

In each test case: 

First line: an integer n(n<=10^5) indicating the size of array

Second line:contain n numbers ai(0<ai<=10000)

 

Output

For each tests: ouput a line contain a number ans.

 

Sample Input

5
1 2 3 4 5

 

Sample Output

23

 

Source

2015 Multi-University Training Contest 1

 

水题，筛法+枚举i

两个数组，L[i]记录离i最近的非互质的j<i，R[i]记录离i最近的非互质的i<j，通过枚举A[i]，ans+=(R[i]-i+1)*(i-L[i]+1)/2 %MOD

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
long long n;
long long d[10005];
long long a[100005];
long long l[100005];
long long r[100005];
#define MOD 1000000007
int main()
{
while(~scanf("%lld",&n))
{
for(long long i = 0;i<n;i++)
scanf("%lld",a+i);
for(long long i = 1;i<=10000;i++)d[i] = -1;
for(long long i = 0;i<n;i++)
{
long long tmp = -1;
for(long long t = 1;t*t<=a[i];t++)
if(a[i]%t==0)
{
tmp = max(tmp,d[t]);
tmp = max(tmp,d[a[i]/t]);
}
tmp = max(tmp,d[a[i]]);
l[i] = tmp+1;
for(long long t = a[i];t<=10000;t+=a[i])
d[t] = i;
}
for(long long i = 1;i<=10000;i++)d[i] = n;
for(long long i = n-1;i>=0;i--)
{
long long tmp = n;
for(long long t = 1;t*t<=a
4000
[i];t++)
if(a[i]%t==0)
{
tmp = min(tmp,d[t]);
tmp = min(tmp,d[a[i]/t]);
}
tmp = min(tmp,d[a[i]]);
r[i] = tmp-1;
for(long long t = a[i];t<=10000;t+=a[i])
d[t] = i;
}
long long ans = 0;
for(long long i = 0;i<n;i++)
{
ans=(ans + (r[i]-i+1)*(i-l[i]+1))%MOD;
// cout<<i<<":"<<l[i]<<" "<<r[i]<<endl;
}
printf("%lld\n",ans);
}
}