题目链接:https://nanti.jisuanke.com/t/A2147
Given a sequence of nnn integers aia_iai.
Let mul(l,r)=∏i=lrai\text{mul}(l, r) = \prod_{i = l}^{r} a_imul(l,r)=∏i=lrai and fac(l,r)\text{fac}(l, r) fac(l,r) be the number of distinct prime factors of mul(l,r)\text{mul}(l, r)mul(l,r).
Please calculate ∑i=1n∑j=infac(i,j)\sum_{i = 1}^{n}\sum_{j = i}^{n}\text{fac}(i, j)∑i=1n∑j=infac(i,j)
Input
The first line contains one integer nnn (1≤n≤1061 \le n \le 10^61≤n≤106) —\text{—}— the length of the sequence.
The second line contains nnn integers aia_iai (1≤i≤n,1≤ai≤1061 \le i \le n, 1 \le a_i \le 10^61≤i≤n,1≤ai≤106) —\text{—}— the sequence.
Output
Print the answer to the equation.
样例输入1
10
99 62 10 47 53 9 83 33 15 24
样例输出1
248
样例输入2
10
6 7 5 5 4 9 9 1 8 12
样例输出2
134
题目大意:任意一个区间的所有数的乘积有多少个不同的素因子,最后计算所有区间的个数的总和;
分析:
/// · ///
参考思路:https://blog.csdn.net/ftx456789/article/details/83116668
这题是计算贡献值,
先筛出每个数的素因子,同时计算每个素因子对后面的数的贡献值,若前面出现过,则从上一个位置开始计算,否则从最左边开始计算
讲下注意点吧
两个int 相乘的时候要long long
~·~
代码:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N=1e6+5;
int prime[N];
unordered_map<int,int> mp;
void get_prime()
{
for(int i=2;i<N;i++)
{
if(!prime[i])
{
prime[++prime[0]]=i;
}
for(int j=1;j<=prime[j]&&i*prime[j]<N;j++)
{
prime[i*prime[j]]=1;
if(i%prime[j]==0)
{
break;
}
}
}
}
int a[N];
int n;
ll ans=0;
void get_ap(int n,int nn,int id)
{
for(int i=1;i<=prime[0]&&prime[i]*prime[i]<=n;i++)
{
if(n%prime[i]==0)
{
while(n%prime[i]==0)
n/=prime[i];
if(mp.count(prime[i]))
{
ans=ans+ (ll)(id-mp[prime[i]])*(nn-id);
mp[prime[i]]=id;
}
else
{
mp[prime[i]]=id;
ans=ans+ (ll)(id+1)*(nn-id);
}
}
}
if(n>1)
{
if(mp.count(n))
{
ans=ans+ (ll)(id-mp[n])*(nn-id);
mp[n]=id;
}
else
{
mp[n]=id;
ans=ans+ (ll)(id+1)*(nn-id);
}
}
return ;
}
int main()
{
get_prime();
while(~scanf("%d",&n)){
ans=0;
mp.clear();
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
get_ap(a[i],n,i);
}
printf("%lld\n",ans);
}
return 0;
}