hdu 5228 OO’s Sequence 多校 思维题

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OO’s Sequence

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

Total Submission(s): 1317 Accepted Submission(s): 467



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

Author

FZUACM



f(l,r)表示l<=i<=r，对任意l<=j<=r,j!=i,i%j!=0，i的个数

求所有f(l,r)之和

开两个数组l[i]和r[i]记录离a[i]最近的是a[i]因数或倍数的下标

由于a[i]小于10000

枚举是a[i]倍数的数就可以

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#define MAXN 111111
#define MOD 1000000007
using namespace std;
int l[MAXN],r[MAXN],pre[MAXN];
int a[MAXN];
int main(){
    int n;
    while(scanf("%d",&n)!=EOF){
        memset(pre,0,sizeof(pre));
        for(int i=1;i<=n;i++){
            l[i]=1;
            r[i]=n;
            scanf("%d",&a[i]);
            for(int j=a[i];j<=10000;j+=a[i])
                if(pre[j]!=0&&r[pre[j]]==n)
                    r[pre[j]]=i-1;
                pre[a[i]]=i;
        }
        memset(pre,0,sizeof(pre));
        for(int i=n;i>=1;i--){
            for(int j=a[i];j<=10000;j+=a[i])
                if(pre[j]!=0&&l[pre[j]]==1)
                    l[pre[j]]=i+1;
                pre[a[i]]=i;
        }
        long long ans=0;
        for(int i=1;i<=n;i++)
            ans=(ans+(i-l[i]+1)*(r[i]-i+1))%MOD;
        printf("%d\n",ans);
    }
    return 0;
}