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 a_i mod a_j=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 a_i(0<a_i<=10000)

 

Output

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

 

Sample Input

51 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 1000000007int 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[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);    }}