C. Multiplicity
time limit per test
3 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given an integer array a1,a2,…,ana1,a2,…,an.
The array bb is called to be a subsequence of aa if it is possible to remove some elements from aa to get bb.
Array b1,b2,…,bkb1,b2,…,bk is called to be good if it is not empty and for every ii (1≤i≤k1≤i≤k) bibi is divisible by ii.
Find the number of good subsequences in aa modulo 109+7109+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 aa has exactly 2n−12n−1 different subsequences (excluding an empty subsequence).
Input
The first line contains an integer nn (1≤n≤1000001≤n≤100000) — the length of the array aa.
The next line contains integers a1,a2,…,ana1,a2,…,an (1≤ai≤1061≤ai≤106).
Output
Print exactly one integer — the number of good subsequences taken modulo 109+7109+7.
Examples
input
Copy
2 1 2
output
Copy
3
input
Copy
5 2 2 1 22 14
output
Copy
13
Note
In the first example, all three non-empty possible subsequences are good: {1}{1}, {1,2}{1,2}, {2}{2}
In the second example, the possible good subsequences are: {2}{2}, {2,2}{2,2}, {2,22}{2,22}, {2,14}{2,14}, {2}{2}, {2,22}{2,22}, {2,14}{2,14}, {1}{1}, {1,22}{1,22}, {1,14}{1,14}, {22}{22}, {22,14}{22,14}, {14}{14}.
Note, that some subsequences are listed more than once, since they occur in the original array multiple times.
【题意】
给定一数组a[]，从a[ ]中除去任意个元素得到b[ ]，求能形成多少“好序列”；好序列的定义是：对于任意的 i 有 b[i]%i == 0（1 ≤ i ≤ size_b[ ]）。
【解题思路】
设dp[i][j]表示前1-i的序列中长度为j的方案数。v[i]表示第i个数的因子，并将因子从小到大排序。易得每个数只有当序列长度刚好达到其拥有的因子时才能构成好序列。
所以dp[i][j]=dp[i-1][j]+dp[i-1][j-1] 当j是a[i]的因子时；否则 dp[i][j]=dp[i-1][j]
所以因子从小到大排序就很重要啦，因为要将二维数组转变为一维的滚动数组，否则会爆内存的哦。
【代码】
#include<bits/stdc++.h>
using namespace std;
const int maxn=1e6+5;
const int mod=1e9+7;
typedef long long LL;
vector<int>v[maxn];
LL dp[maxn],a[maxn];
int main()
{
LL n;
scanf("%lld",&n);
for(int i=1;i<=n;i++)scanf("%lld",&a[i]);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=sqrt(a[i]);j++)
{
if(a[i]%j==0)
{
v[i].push_back(j);
if(a[i]!=j*j)v[i].push_back(a[i]/j);//例如7*8，避免8没有存入
}
}
sort(v[i].begin(),v[i].end());
}
dp[0]=1;
for(int i=1;i<=n;i++)
{
for(int j=v[i].size()-1;j>=0;j--)
{
dp[v[i][j]]+=dp[v[i][j]-1];
dp[v[i][j]]%=mod;
}
}
LL ans=0;
for(int i=1;i<=n;i++)ans+=dp[i];
printf("%lld\n",ans%mod);
}