題目鏈接：hdu 4961 Boring Sum

題目大意：給定ai數組;

• 構造bi,?k=max(j|0<j<i,aj%ai=0),?bi=ak;
• 構造ci,?k=min(j|i<j≤n,aj%ai=0),?ci=ak;
  求∑i=1nbi?ci

解題思路：由于ai≤105,所以預先處理好每一個數的因子，然后在處理bi，ci數組的時候，每次遍歷一個數。就將其全部的因子更新，對于bi維護最大值，對于ci維護最小值。

#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; }

轉載于:https://www.cnblogs.com/lytwajue/p/6715549.html

總結

以上是生活随笔為你收集整理的hdu 4961 Boring Sum(高效)的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。