2018南京区域赛——G

https://blog.csdn.net/m0_37624640/article/details/83276324

这篇博客很详细，也很容易理解。求每个的贡献贡献，得到ans[i​]=(n−prime[i][pos]+1)∗(prime[i][pos]−prime[i][pos−1]);

#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<cstdio>
#include<stack>
#include<cmath>
#include<iostream>
#define ll long long
#define lowbit(x) x&(-x)
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=1e6+7;
int n,a[maxn];
vector<int> prime,pos[maxn];
bool vis[maxn]={false};
void dive(int x,int p)
{
    for(int i=2;i<=sqrt(x);i++)
    {
        if(x%i==0)
        {
            pos[i].push_back(p);   
            if(!vis[i])
            {   
                prime.push_back(i);
                vis[i]=true;
            }
            while(x%i==0)
                x/=i;
        }
    }
    if(x>1)
    {
        pos[x].push_back(p);
        if(!vis[x])
        {
            prime.push_back(x);
            vis[x]=true;
        }
    }
}
int main()
{
    for(int i=0;i<maxn;i++) 
        pos[i].push_back(0);
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        dive(a[i],i);      
    }
    ll ans=0;
    for(int i=0;i<prime.size();i++)
    {
        for(int j=1;j<pos[prime[i]].size();j++)
        {
            ans+=(ll)(pos[prime[i]][j]-pos[prime[i]][j-1])*(ll)(n-pos[prime[i]][j]+1);
        }
    }
    printf("%lld
",ans);
    return 0;
}