C. Multiplicity

題目連線
You are given an integer array a1,a2,…,an

.

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 b1,b2,…,bk
is called to be good if it is not empty and for every i (1≤i≤k) bi is divisible by i
Find the number of good subsequences in a
modulo 109+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 2n−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 a1,a2,…,an
(1≤ai≤106).
Output
Print exactly one integer — the number of good subsequences taken modulo 109+7
題目大意：給你一個數組a,從中可以得到多少個子序列b,並且序列b中的bi能被i整除。首先想到的是二維陣列dp[i][j],i表示在a陣列的前i個數，j表示陣列b的長度。那麼dp[i][j]表示的就是在前i個數中陣列b長度為j的個數。初始化中dp[0][0]到dp[n][0]初始化為1；
如果（a[i]%j==0）
dp[i][j]=dp[i-1][j-1]+dp[i-1][j]；否則dp[i][j]=dp[i-1][j];由於該題資料比較大，所以用二維的話不行，由於每次更新都是從上一位置得到所以可以把二維滾動成一維dp[i]中的i表示陣列b的長度。每次更新只能將因數從大到小進行更新，否則會的話當前的更新會受到當前更新數的影響，自己手動進行模擬後就會發現。

#include<stdio.h>
#include<math.h>
#include<map>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
map<int,int>mp;
ll dp[1000004]= {0},ans=0;
int main()
{

    int n,i,
 
j,a;
    dp[0]=1;
    scanf("%d",&n);
    for(i=1; i<=n; i++)
    {
        vector<ll>v;
        scanf("%d",&a);
        for(j=1; j*j<=a; j++)
        {
            if(a%j==0)
            {
                v.push_back(j);
                if(a/j!=j)
                    v.push_back(a/j);

            }
        }
        sort(v.begin(),v.end());
        reverse(v.begin(),v.end());

        for(int k=0;k<v.size();k++)
        {
            dp[v[k]]+=dp[v[k]-1];
            dp[v[k]]%=mod;
        }
    }
    for(i=1;i<=n;i++)
        ans+=dp[i];
    ans%=mod;
    printf("%lld\n",ans);
    return 0;
}

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