cf luogu 暴力是挨个考虑,这里不妨整体考虑,先把所有T恤按照价值从大到小为第一关键字,价格从小到大为第二关键字排序,然后依次考虑,可以发现这时候剩余钱数\(>c_i\)的会买,这会导致他们钱数\(-c_i\),答案\(+1\).因为这一段人是连续区间,所以考虑用平衡树维护,每次找到钱数\(.=c_i\)的点,打上减钱数和答案\(+1\)的标记,再和其他点合并成新平衡树 不过直接这样做显然是

题意： 有 n n n种 T 恤，每种有价格 c i c_i ci​ 和品质 q i q_i qi​。 有 m m m 个人要买 T 恤，第 i i i 个人有 v i v_i vi​​ 元，每人每次都会买一件能买得起的 q i q_i qi​​ 最大的 T 恤。一个人只能买一种 T 恤一件，所有人之间都是独立的。 问最后每个人买了多少件 T 恤？如果有多个 q i q_i qi​​ 最大的 T

题目： Codehorses has just hosted the second Codehorses Cup. This year, the same as the previous one, organizers are giving T-shirts for the winners. The valid sizes of T-shirts are either "M" or from 00

C. Kuroni and Impossible Calculation time limit per test1 second memory limit per test256 megabytes inputstandard input outputstandard output To become the king of Codeforces, Kuroni has to solve the

D. T-shirts Distribution time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output The organizers of a programming contest have decided to present t-

A. Codehorses T-shirts Codehorses has just hosted the second Codehorses Cup. This year, the same as the previous one, organizers are giving T-shirts for the winners. The valid sizes of T-shirts are ei

Working in a boutique folding and putting in order T-shirts according to their sizes seems very easy. But is it really so simple?     Given n objects of different sizes, how many different arrangement

Codehorses has just hosted the second Codehorses Cup. This year, the same as the previous one, organizers are giving T-shirts for the winners. The valid sizes of T-shirts are either “M” or from 0 to 3

Problem D. T-shirts 来源：ACM International Collegiate Programming Contest, Syrian Collegiate Programming Contest (2014) Syria, September, 23, 2014  CF链接Gym 100500D Problem Description It was the second

map 由{键，值}对组成的集合，内部实现是一颗红黑树 好啦，具体自行百度。 说起来实在惭愧，map我并不会，只会一个操作就是赋值，还只是一种赋值方法 题目链接：点我跳转 Codehorses has just hosted the second Codehorses Cup. This year, the same as the previous one, organizers are givi

A. Codehorses T-shirts time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Codehorses has just hosted the second Codehorses Cup. This year, th

A题： 既然保证了不同那么排个序就行了。 #include <bits/stdc++.h> using namespace std; int a[105],b[105]; int main() { int t; scanf("%d",&t); while(t--){ int n; scanf("%d",&n); for(int i=1;i<=n;i+

描述 (Description) 字符\t匹配制表符。 例子 (Example) 以下示例显示了字符匹配的用法。 package com.wenjiangs; import java.util.regex.Matcher; import java.util.regex.Pattern; public class CharactersDemo { private static final St

我只是想说清楚，我的意思是这样的- 另外，如果我访问第一个数组以外的元素，也会遇到同样的问题，即(INT*)arr+13。它会属于越界访问的条款吗？因为我是在第一个数组的边界之外访问的。

描述 (Description) 如果存在这样的注释，则java.lang.reflect.Method.getAnnotation(Class《T》 annotationClass)方法返回指定类型的此元素的注释，否则为null。 声明 (Declaration) 以下是java.lang.reflect.Method.getAnnotation(Class《T》 annotationClass

描述 (Description) 如果存在这样的注释，则java.lang.reflect.Field.getAnnotation(Class《T》 annotationClass)方法返回指定类型的此元素的注释，否则为null。 声明 (Declaration) 以下是java.lang.reflect.Field.getAnnotation(Class《T》 annotationClass)方

描述 (Description) 如果存在这样的注释，则java.lang.reflect.Constructor.getAnnotation(Class《T》 annotationClass)方法返回指定类型的此元素的注释，否则为null。 声明 (Declaration) 以下是java.lang.reflect.Constructor.getAnnotation(Class《T》 annot

描述 (Description) java.lang.reflect.AccessibleObject.getAnnotation(Class annotationClass) java.lang.reflect.AccessibleObject.getAnnotation(Class annotationClass) 如果存在这样的注释，则method返回此元素的指定类型的注释，否则返回null