hdu 5288 OO’s Sequence 2015 Multi-University Training Contest 1

OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 83    Accepted Submission(s): 34

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

求一个每个区间中，满足其他数字%该数字 ！= 0的数字个数，把每个区间满足条件的个数相加

方法：

枚举每个位置，求包含这个数字的区间个数，做加法。

对于ai，看ai左边最近的位置l，al是ai的因数，右边位置r，ar是ai的因数。那么满足的个数就是

（i-l)*(r-i)

预处理所有1-10000的因数。

从左到右，用pre[i]表示数字i出现的i的最右边的位置。算出每个位置的l。

同理从右到左，算出每个位置的r。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<vector>using namespace std;#define maxn 100007vector<int> head[10007];int pre[10007];int num[maxn];int lef[maxn];int righ[maxn];int main(){    int n;    for(int i = 1;i <= 10000;i++){        head[i].clear();        for(int j = 1;j*j<=i;j++){            if(i%j == 0){                head[i].push_back(j);                head[i].push_back(i/j);            }        }    }    while(scanf("%d",&n)!=EOF){        for(int i = 1;i <= n; i++)            scanf("%d",&num[i]);        memset(pre,0,sizeof(pre));        for(int i = 1;i <= n; i++){            int u = num[i];            int p = 0;            for(int j = 0;j < head[u].size(); j++)                p = max(p,pre[head[u][j]]);            lef[i] = p;            pre[u] = i;        }        memset(pre,0x3f,sizeof(pre));        for(int i = n; i > 0; i--){            int u = num[i];            int p = n+1;            for(int j = 0;j < head[u].size();j++)                p = min(p,pre[head[u][j]]);            righ[i] = p;            pre[u] = i;        }        long long ans = 0,l,r;        long long mod = 1000000007;        for(int i = 1; i <= n; i++){            ans += (long long)(i-lef[i])*(righ[i]-i);            ans %= mod;        }        cout<<ans<<endl;    }    return 0;}