Discription
Number theory is interesting, while this problem is boring.
Here is the problem. Given an integer sequence a 1, a 2, …, a n, let S(i) = {j|1<=j<i, and a j is a multiple of a i}. If S(i) is not empty, let f(i) be the maximum integer in S(i); otherwise, f(i) = i. Now we define b i as a f(i). Similarly, let T(i) = {j|i<j<=n, and a j is a multiple of a i}. If T(i) is not empty, let g(i) be the minimum integer in T(i); otherwise, g(i) = i. Now we define c i as a g(i). The boring sum of this sequence is defined as b 1 * c 1 + b 2 * c 2 + … + b n * c n.
Given an integer sequence, your task is to calculate its boring sum.
Here is the problem. Given an integer sequence a 1, a 2, …, a n, let S(i) = {j|1<=j<i, and a j is a multiple of a i}. If S(i) is not empty, let f(i) be the maximum integer in S(i); otherwise, f(i) = i. Now we define b i as a f(i). Similarly, let T(i) = {j|i<j<=n, and a j is a multiple of a i}. If T(i) is not empty, let g(i) be the minimum integer in T(i); otherwise, g(i) = i. Now we define c i as a g(i). The boring sum of this sequence is defined as b 1 * c 1 + b 2 * c 2 + … + b n * c n.
Given an integer sequence, your task is to calculate its boring sum.
Input
The input contains multiple test cases.
Each case consists of two lines. The first line contains an integer n (1<=n<=100000). The second line contains n integers a 1, a 2, …, a n (1<= ai<=100000).
The input is terminated by n = 0.
Output
Output the answer in a line.
Sample Input
5 1 4 2 3 9 0
Sample Output
136
Hint
In the sample, b1=1, c1=4, b2=4, c2=4, b3=4, c3=2, b4=3, c4=9, b5=9, c5=9, so b1 * c1 + b2 * c2 + … + b5 * c5 = 136.
预处理一下每个数的约数，直接暴力做就行了。
#include<iostream> #include<cstdio> #include<cstdlib> #include<cmath> #include<algorithm> #include<vector> #include<cstring> #define ll long long #define maxn 100005 #define pb push_back using namespace std; ll tot=0; vector<int> son[maxn]; int n,m,a[maxn],f[maxn]; int mult[maxn],g[maxn],to; inline void init(){ for(int i=1;i<=100000;i++) for(int j=i;j<=100000;j+=i) son[j].pb(i); } int main(){ init(); while(scanf("%d",&n)==1&&n){ memset(mult,0,sizeof(mult)); for(int i=1;i<=n;i++){ scanf("%d",a+i); f[i]=mult[a[i]]; if(!f[i]) f[i]=i; for(int j=son[a[i]].size()-1;j>=0;j--){ to=son[a[i]][j]; mult[to]=max(mult[to],i); } } memset(mult,0x3f,sizeof(mult)); for(int i=n;i;i--){ g[i]=mult[a[i]]; if(g[i]==mult[0]) g[i]=i; for(int j=son[a[i]].size()-1;j>=0;j--){ to=son[a[i]][j]; mult[to]=min(mult[to],i); } } tot=0; for(int i=1;i<=n;i++) tot+=(ll)a[f[i]]*(ll)a[g[i]]; printf("%lld\n",tot); } return 0; }
