题意:给定 n 个物品,然后有 m 个人买东西,他们有 x 元钱,然后从 l - r 这个区间内买东西,对于每个物品都尽可能多的买,问你最少剩下多少钱. 析:对于物品,尽可能多的买的意思就是对这个物品价格取模,但是对于价格比我的钱还多,那么就没有意义,对取模比我的钱少的,那取模至少减少一半,所以最多只要60多次就可以结束,为了快速找到第一个比我的钱少的,使用线段树. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000"…
题意:n个数,m次询问,每次给你一个询问v,l,r,问你v%a[l]%a[l+1]%...%a[r]是多少. a%b,结果要么不变,要么至少缩小到a的一半,于是用线段树,每次询问当前区间最靠左侧的小于等于当前数的值是多少,只需不超过log次询问就能使该数模完,就行了. O(n(logn)^2). #include<cstdio> #include<algorithm> using namespace std; typedef long long ll; int n,m; ll mi…
题目链接:https://codeforces.com/gym/101982/attachments 要你求覆盖奇数次的矩形面积并,每次更新时减去原先的值即可实现奇数次有效,下推时为保证线段长度不变左儿子的值为x[mid]-x[l]再减原来的值,右儿子的值为x[r]-x[mid]再减原来的值 #include<iostream> #include<algorithm> using namespace std; #define ll long long #define maxn 20…
There are nn people at the round gaming table. Each of them has a set of cards. Every card contains some number xx. Players make turns consecutively, one after another, starting from the player number 1. A player in his turn can either skip his turn…
传送门 题意:给一棵带颜色的树,可以给子树染色或者问子树里有几种不同的颜色,颜色值不超过606060. 思路:颜色值很小,因此状压一个区间里的颜色用线段树取并集即可. 代码: #include<bits/stdc++.h> #define ri register int using namespace std; inline int read(){ int ans=0; char ch=getchar(); while(!isdigit(ch))ch=getchar(); while(isdi…
D. The Child and Sequence At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks. Fortunately, Picks remembers how t…