题目链接：hdu 4961 Boring Sum
题目大意：给定ai数组;
- 构造bi, k=max(j|0<j<i,aj%ai=0), bi=ak;
- 构造ci, k=min(j|i<j≤n,aj%ai=0), ci=ak;
求∑i=1nbi∗ci
解题思路：由于ai≤105,所以预先处理好每一个数的因子，然后在处理bi，ci数组的时候，每次遍历一个数。就将其全部的因子更新，对于bi维护最大值，对于ci维护最小值。
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>using namespace std;
typedef long long ll;
const int maxn = 1e5;
const int INF = 0x3f3f3f3f;int n, arr[maxn+5], b[maxn+5], c[maxn+5], v[maxn+5];
vector<int> g[maxn+5];void get_factor (int n) {for (int i = 1; i <= n; i++)g[i].clear();for (int i = 1; i <= n; i++) {for (int j = i; j <= n; j += i)g[j].push_back(i);}
}void init () {memset(v, 0, sizeof(v));for (int i = 1; i <= n; i++) {int u = arr[i];int k = (v[u] == 0 ? i :v[u]);b[i] = arr[k];for (int j = 0; j < g[u].size(); j++)v[g[u][j]] = max(v[g[u][j]], i);}memset(v, INF, sizeof(v));for (int i = n; i >= 1; i--) {int u = arr[i];int k = (v[u] == INF ? i : v[u]);c[i] = arr[k];for (int j = 0; j < g[u].size(); j++)v[g[u][j]] = min(v[g[u][j]], i);}
}int main () {get_factor(maxn);while (scanf("%d", &n) == 1 && n) {for (int i = 1; i <= n; i++)scanf("%d", &arr[i]);init();ll ans = 0;for (int i = 1; i <= n; i++)ans = ans + b[i] * 1LL * c[i];printf("%I64d\n", ans);}return 0;
}
