Feel Good   Description Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated to studying how good or bad days influent people's memories about some period of life. A new idea Bill has recently devel…

Feel Good Time Limit: 3000MS   Memory Limit: Unknown   64bit IO Format: %lld & %llu Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated to studying how good or bad days influent people's memories a…

题目: 思路: 预处理出a[i]在哪个范围区间内是最小的,然后直接遍历a数组求答案就可以了. 这个预处理的技巧巧妙的用了之前的处理结果.(大佬tql) 代码: #include <bits/stdc++.h> #define inf 0x3f3f3f3f #define MAX 1e3 #define FRE() freopen("in.txt","r",stdin) #define FRO() freopen("out.txt",&…

Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated to studying how good or bad days influent people's memories about some period of life. A new idea Bill has recently developed assigns a non-negat…

Feel Good Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 16786   Accepted: 4627 Case Time Limit: 1000MS   Special Judge Description Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated…

1619 - Feel Good Time limit: 3.000 seconds   Bill is developing a new mathematical theory for human emotions. His recent investigations are dedi- cated to studying how good or bad days in uent people's memories about some period of life. A new idea B…

#include <bits/stdc++.h> using namespace std; const int maxn = 1000010; struct node { int num; int pos; }; int main() { // freopen("input", "r", stdin); int n; int a[maxn], s[maxn]; while (scanf("%d", &n) != EOF) {…

Feel Good Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 12409   Accepted: 3484 Case Time Limit: 1000MS   Special Judge Description Bill is developing a new mathematical theory for human emotions. His recent investigations are dedicated…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 用单调队列求出l[i]和r[i] 分别表示i的左边最近的大于a[i]的数的位置以及i右边最近的大于a[i]的数的位置. 则l[i]+1..r[i]-1就是a[i]这个数作为最小数的最大管辖区间了. 写个前缀和就好. 然后取a[i]*区间l[i]+1..r[i]-1的和 中的最大值. 并不是special judge 多个相同区间. 取区间长度最短的区间. 如果仍有多个答案. 取左端点最小的那个区间. (全都是0的情况要注意,直接…

这道题貌似可以用滑动窗口或者单调栈做, 但是我都没有用到. 这道题要求连续子序列中和乘上最小值最大, 那么我们就可以求出每一个元素, 以它为最小值的的最大区间的值, 然后取max就ok了.那么怎么求呢? 我可以初始化出一个第一个小于当前元素的的元素的位置, 也就是说初始化出以当前元素为最小值的 左右端点, 求出之后枚举一遍就ok了. 那么关键是怎么求呢? 这里有一点扫描法的味道. 假设当前序列为1243 以求右端点为例, 右端点的位置也就是在右侧第一个小于当前元素的元素的位置. 那么如果序列是递…