2015 Multi-University Training Contest 1 OO’s Sequence

OO’s Sequence

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 312    Accepted Submission(s): 107

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

\[ \sum_{i=1}^{n}\sum_{j=i}^{n}f(i,j)mod(10^{9}+7) \]

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

23

 

 

Source

2015 Multi-University Training Contest 1 

 

 解题：

 

需要先使用$\sqrt{n}$的时间复杂度把每个数的倍数所在的下标找到

然后我们对每个数，如果她的倍数所在的下标比它大，那么当前这个数是它的倍数左边的约数，求最大的左下标，

如果它的倍数所在的下标比它小，那么当前这个数是它的倍数右边的约数，求最小的右下标

 

$(i-L[i])\times (R[i] - i)$就是i所在的区间的个数，这个区间没有它的约数

 

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 10010;
const LL mod = 1000000007;
vector<int>g[maxn],factor[maxn];
int d[100010];
LL L[100010],R[100010],n;
int main() {
    while(~scanf("%d",&n)) {
        for(int i = 0; i < maxn; ++i) {
            g[i].clear();
            factor[i].clear();
        }
        for(int i = 1; i <= n; ++i) {
            scanf("%d",d+i);
            g[d[i]].push_back(i);
            L[i] = 0;
            R[i] = n+1;
        }
        for(int i = 1; i <= n; ++i) {
            for(int j = 1,m = sqrt(d[i]); j <= m; ++j) {
                if(d[i]%j) continue;
                int tmp = d[i]/j;
                if(g[tmp].size()) factor[tmp].push_back(i);
                if(tmp == j) continue;
                if(g[j].size()) factor[j].push_back(i);
            }
        }
        for(int i = 1; i <= n; ++i) {
            for(int j = factor[d[i]].size()-1; j >= 0; --j) {
                int now = factor[d[i]][j];
                if(now < i) R[now] = min(R[now],(LL)i);
                if(now > i) L[now] = max(L[now],(LL)i);
            }
        }
        LL ret = 0;
        for(int i = 1; i <= n; ++i)
            ret = (ret + (i - L[i])*(R[i] - i)%mod)%mod;
        cout<<ret<<endl;
    }
    return 0;
}