[codeforces1061C]Multiplicity

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C. Multiplicity
time limit per test : 3 seconds
memory limit per test : 256 megabytes

You are given an integer array $a_1,a_2,…,a_n$.
The array $b$ is called to be a subsequence of a if it is possible to remove some elements from a to get $b$.Array $b_1,b_2,…,b_k$ is called to be good if it is not empty and for every $i (1≤i≤k) b_i$ is divisible by $i$. Find the number of good subsequences in amodulo $10^9+7$.

Two subsequences are considered different if index sets of numbers included in them are different. That is, the values ​of the elements ​do not matter in the comparison of subsequences. In particular, the array $a$ has exactly $2^n−1$ different subsequences (excluding an empty subsequence).

Input

The first line contains an integer $n(1≤n≤100000)$ — the length of the array $a$.
The next line contains integers $a_1,a_2,…,a_n(1≤a_i≤10^6)$.
Output
Print exactly one integer — the number of good subsequences taken modulo $10^9+7$
Examples
Input

2
1 2

Output

3

Input

5
2 2 1 22 14

Output

13

Note

In the first example, all three non-empty possible subsequences are good: { $1$} , { $1,2$}, { $2$}.
In the second example, the possible good subsequences are: { $2$} , { $2,2$}, { $2,22$}, {2,14}, {2}, {2,22}, {2,14}, {1}, {1,22}, {1,14}, {22}, {22,14}, {14}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times.

题意：
给定一个数组{ $a_k$}，问这个数组的所有子序列{ $b_k$}中，有多少子序列满足:对于所有的 $i(1<=i<=k)$满足 $b_i$ 是 $i$ 的倍数，答案对 $10^9+7$取模

题解：
先处理出 $a[i]$的所有因数。
接着就是DP
$f[i][j]$表示从 $1$~ $i$前面所有序列中，长度为 $j$的方案数
所以 $f[i][j]=(f[i][j]+f[i][j-1])$% $MOD$
$j$可以通过枚举 $a[i]$的因数得到
然后我们发现可以把 $i$这个维度压掉…

#include<bits/stdc++.h>
#define LiangJiaJun main
#define ll long long
#define MOD 1000000007
using namespace std;
vector<int>v[100004];
int n,f[1000004];
int w33ha(){
    memset(f,0,sizeof(f));
    for(int i=1;i<=n;i++){
        int x;
        scanf("%d",&x);
        v[i].clear();
        for(int j=1;j<=sqrt(x);j++){
            if(x%j==0){
                v[i].push_back(j);
                if(x!=j*j)v[i].push_back(x/j);
            }
        }
        sort(v[i].begin(),v[i].end());
    }
    ll ans=0;
    f[0]=1;
    for(int i=1;i<=n;i++){
        for(int j=v[i].size()-1;j>=0;j--){
            f[v[i][j]]=(f[v[i][j]]+f[v[i][j]-1])%MOD;
            ans=(ans+f[v[i][j]-1])%MOD;
        }
    }
    printf("%d\n",ans);
    return 0;
}
int LiangJiaJun(){
    while(scanf("%d",&n)!=EOF)w33ha();
    return 0;
}