HDU-4961 Boring Sum （模拟）
Boring Sum
http://acm.hdu.edu.cn/showproblem.php?pid=4961
Problem Description
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<= a i<=100000).
The input is terminated by n = 0.
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<= a i<=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
136HintIn 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.
又是无限接近正解而没有写出来，原本想着维护一个[1,i)中是a[i]点倍数点下标点最值，但是两个最值都要维护，就不会了。。。
先正着扫一遍，维护[1,i)中是a[i]点倍数点下标点最大值，并及时更新b[i]；再倒着扫一遍，维护(i,n]中是a[i]点倍数点下标点最小值，并及时更新c[i]
最后相乘得出答案
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;
const int MAXN=100005;
int n,f,g;
int a[MAXN],b[MAXN],c[MAXN],indx[MAXN];
long long ans;
vector<int> factor[MAXN];//factor[i]表示i点因子的表
int main() {
for(int j=1;j<=MAXN;++j)
for(int i=j;i<=MAXN;i+=j)
factor[i].push_back(j);
while(scanf("%d",&n),n!=0) {
memset(indx,0,sizeof(indx));
for(int i=1;i<=n;++i) {//indx[x]表示[1,i)中是a[i]的倍数的最大下标
scanf("%d",a+i);
f=indx[a[i]]==0?i:indx[a[i]];
b[i]=a[f];
for(int j=0;j<factor[a[i]].size();++j)
indx[factor[a[i]][j]]=max(indx[factor[a[i]][j]],i);
}
memset(indx,0x3f,sizeof(indx));
for(int i=n;i>0;--i) {//indx[x]表示(i,n]中是a[i]的倍数的最小下标
g=indx[a[i]]==0x3f3f3f3f?i:indx[a[i]];
c[i]=a[g];
for(int j=0;j<factor[a[i]].size();++j)
indx[factor[a[i]][j]]=min(indx[factor[a[i]][j]],i);
}
ans=0;
for(int i=1;i<=n;++i)
ans+=(long long)b[i]*c[i];
printf("%I64d\n",ans);
}
return 0;
}