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…
ACM-ICPC 2018 徐州赛区网络预赛 G. Trace (思维,贪心) Trace 问答问题反馈 只看题面 35.78% 1000ms 262144K There's a beach in the first quadrant. And from time to time, there are sea waves. A wave ( xx , yy ) means the wave is a rectangle whose vertexes are ( 00 , 00 ), ( xx ,…
ACM-ICPC 2018 徐州赛区网络预赛 J. Maze Designer J. Maze Designer After the long vacation, the maze designer master has to do his job. A tour company gives him a map which is a rectangle. The map consists of N \times MN×M little squares. That is to say, the h…
传送门 题面: In a world where ordinary people cannot reach, a boy named "Koutarou" and a girl named "Sena" are playing a video game. The game system of this video game is quite unique: in the process of playing this game, you need to consta…
In a world where ordinary people cannot reach, a boy named "Koutarou" and a girl named "Sena" are playing a video game. The game system of this video game is quite unique: in the process of playing this game, you need to constantly fac…
262144K 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 <x, y>. If xi= xj and yi…
Mur loves hash algorithm, and he sometimes encrypt another one's name, and call him with that encrypted value. For instance, he calls Kimura KMR, and calls Suzuki YJSNPI. One day he read a book about SHA-256,which can transit a string into just 256 b…
A. Hard to prepare 题意:有n个客人做成一圈,有$2^k$种面具,对于每种面具有一种面具不能使相邻的两个人戴,共有多少种做法. 思路: 把题意转化成相邻的人不能带同种面具.总数为$(2^k)^n$,减去一对相邻的客人戴同种面具$(2^k)^{(n-1)}*C(n,1)$,其中重复了两对相邻的客人戴同种面具$(2^k)^{(n-2)}*C(n,2)$,依次容斥. 最后所有人都戴同种面具的情况额外考虑,当n是奇数时,n-1对客人相同即所有人相同.n为偶数时,n-1对客人相同时用公式…
A. Hard to prepare #include <bits/stdc++.h> using namespace std; ; ]; ]; int main() { int T; int n,k; bit[] = ; ; i <= ; i++) { bit[i] = *bit[i-]%MOD; } scanf("%d", &T); while (T--) { scanf("%d%d", &n, &k); f[] = b…
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…
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…
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…
I. query 题目链接: Problem Description Given a permutation \(p\) of length \(n\), you are asked to answer \(m\) queries, each query can be represented as a pair \((l ,r )\), you need to find the number of pair \((i ,j)\) such that \(l \le i < j \le r\) a…
