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

最新推荐文章于 2018-07-16 18:00:13 发布

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#素数 #晒法 #多校

本文详细介绍了如何解决OO序列中关于非互质数的复杂问题，通过筛法与枚举技巧，实现高效计算，特别适合大规模数据集的处理。

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

\sum i = 1 n \sum j = i n f (i, j) m o d （ 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

  
   5
1 2 3 4 5

Sample Output

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[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);
    }
}