H.Ryuji doesn't want to study 27.34% 1000ms 262144K   Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i]a[i]. Unfortunately, the longer he learns, the fewer he gets. That…

262144K   Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i]a[i]. Unfortunately, the longer he learns, the fewer he gets. That means, if he reads books from ll to rr, he…

Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge a[i]a[i]. Unfortunately, the longer he learns, the fewer he gets. That means, if he reads books from ll to rr, he will get a…

F. Features Track Morgana is learning computer vision, and he likes cats, too. One day he wants to find the cat movement from a cat video. To do this, he extracts cat features in each frame. A cat feature is a two-dimension vector <xx, yy>. If x_ixi…

https://nanti.jisuanke.com/t/31460 题意 两个操作.1:查询区间[l,r]的和,设长度为L=r-l+1, sum=a[l]*L+a[l+1]*(L-1)+...+a[r].2:将第a个位置修改为b. 分析 变形一下,sum=a[l]*(r-l+1)+a[l+1]*(r-(l+1)-1)+...+a[r](r-r+1)=(r+1)*a[l]-a[l]*l.因此,可维护两个树状数组计算.至于更新操作,实质就是看变化前后的差值,变大就相当与加上差值,否则就是减.注意用…

题意:给你数组a,有两个操作 1 l r,计算l到r的答案:a[l]×L+a[l+1]×(L−1)+⋯+a[r−1]×2+a[r] (L is the length of [ l, r ] that equals to r - l + 1),或者 2 i b:把第i个换成b 思路:用一个树状数组存i的前缀和,再用一个树状数组存(n - i + 1)*a[ i ]的前缀和,这样算出后面那个的区间差减去前一个的区间差的某个倍数就会成为答案. 代码: #include<queue> #include…

Output For each question, output one line with one integer represent the answer. 样例输入 5 3 1 2 3 4 5 1 1 3 2 5 0 1 4 5 样例输出 10 8两个树状数组一个维护a[i]前缀合,一个维护(n-i+1)*a[i]前缀和. #include <iostream> #include <algorithm> #include <cstring> #include &l…

题目链接:https://nanti.jisuanke.com/t/31460 Ryuji is not a good student, and he doesn't want to study. But there are n books he should learn, each book has its knowledge $a[i]$. Unfortunately, the longer he learns, the fewer he gets. That means, if he re…

题意: 1.区间求 a[l]×L+a[l+1]×(L−1)+⋯+a[r−1]×2+a[r](L is the length of [ l, r ] that equals to r - l + 1) 2.单点修改 解析: 只有这两个操作 并且区间求得值与前缀和极为相似 所以嘛 就要想到树状数组 树状数组能够维护前缀和 所以区间 [ l, r ] 就是r的前缀和减去l的前缀和 但每个a[i]的系数都不同 求前缀和无用..所以要想到化简一下  把每个a[i]的系数都化为一样的或者定量 由上面的式子化…

简单数学变换+线段树 简单数据结构签到题不解释 本来应该贴板子的,鉴于最近写代码太少了,而且由于要用两个线段树,平时板子都是一个的.以及板子在队友那.就当熟悉写代码,自己写了一下. #include <bits/stdc++.h> using namespace std; #define dual(i, n) (n) + 1 - (i) typedef long long ll; int n, q; const int maxn = 1e+5 + 5; typedef ll arr[maxn]…