HDU 5288 OO’s Sequence

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原文链接：http://www.cnblogs.com/Basasuya/p/8433773.html

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#java

本文介绍了一种利用线性筛法计算数组中任意元素与其他元素互质的区间个数的算法，并通过实例展示了如何实现并优化该算法。

OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2927 Accepted Submission(s): 1041

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

/*
∑i=1-n ∑j=i-n f(i,j) 
其实是在求对于任一个i满足与其他成员互质的区间个数
则就可以采用类似线性筛的方法，将i的左右端点给记录下来
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
typedef long long LL;
const int mod=1000000007;
#define N 100005
int a[N];
int Hasha[10005],Hashb[10005];
int positiona[N],positionb[N];
int main(){
    int x;
    int sum;
    int i,j,k;
    while(~scanf("%d",&x)){
        sum=0;
        memset(Hasha,0,sizeof(Hasha));
        for(i=1;i<=10000;i++) Hashb[i]=x+1;
        for(i=1;i<=x;i++) scanf("%d",&a[i]);
        for(i=1;i<=x;i++){
            positiona[i]=Hasha[a[i]];
            for(j=a[i];j<10005;j+=a[i]) Hasha[j]=i;
        }
        for(i=x;i>=1;i--){
            positionb[i]=Hashb[a[i]]-1;
            for(j=a[i];j<10005;j+=a[i]) Hashb[j]=i;
        }
        for(i=1;i<=x;i++){
            //printf("%lld %lld\n",positiona[i],positionb[i]);
            sum=(sum+(i-positiona[i])*(positionb[i]+1-i))%mod;
        }
        printf("%d\n",sum);
    }
    return 0;
}

转载于:https://www.cnblogs.com/Basasuya/p/8433773.html