hdu 5288 OO’s Sequence 2015 Multi-University Training Contest 1-优快云博客

本文介绍了一种方法，用于计算数组中每个区间的满足特定数学条件的数字个数，并将这些个数进行累加。通过枚举每个位置并查找其左右边界，利用预处理的因数信息来优化计算过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

\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

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 100007
vector<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;
}