签到题目,排序然后按序清理掉一定会输的结果就可以. ac代码: #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; ]; int main() { int t; cin>>t; while(t--) { int n,k; ; scanf("%d %d",&n,&k);…

Rikka with Competition 给出一个大小为n的集合\(\{a_i\}\),每次从集合中随机挑出一对数\(a_i,a_j\),如果\(|a_i-a_j|>K\),那么从集合中删掉较小的那个数字,否则随机删掉一个数字,显然会有一个数字保留到最后,现在询问能够有可能保留到最后一个数字的数字个数,\(n\leq 10^5\). 解 序列无序,不妨排个序构成一个序列,显然序列最后一个元素必然有机会留到最后,此时考虑倒数第二个元素,如果它和最后一个元素的差的绝对值不超过K那么也有机会保留到…

Problem Description As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: A wrestling match will be held tomorrow. n players will take part in it. The ith player…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=6095 题目大意: 任意两个人相比,相差大于K,分低的淘汰,否则两人都有可能赢,剩下的继续比,问有最多多少人可能赢? 思路: 排序,如果Ai - Ai-1的值大于k,那从Ai-1开始的人都不可能有机会赢了. #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #inc…

2017-09-22 22:01:19 writer:pprp As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: A wrestling match will be held tomorrow. nn players will take part in it. T…

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has nn positive A1−AnA1−An and their sum is mm. Then for each subset SS of AA, Yuta calculates the sum…

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: Yuta has a non-direct graph with n vertices and n+1 edges. Rikka can choose some of the edges (at least one…

---恢复内容开始--- 题目链接: https://vjudge.net/problem/1812693/origin 这题的mod运算很恶心,真的... 本题有正反两个思路,一个是正面求解其不能成立的情况, 一个是反面求解,用total减. 我用的是正面求解. 一共有种情况: 1. 全是球 :poww(2, a)*poww(2,c)-1 2. 全是拍 : poww(2, a)*poww(2, b)-1 3. 一拍多球(2种可能):    1 -- (poww(2, a)*poww(2, c)…

题意:给定$n$个数,代表$n$个选手的能量高低,现在再给一个$k$,任意在$n$个选手中挑取两个选手比赛,如果$|a_i−a_j|>K$,那么能量高的选手获胜,另一个将被淘汰,否则两个人都有机会获胜,现在要你求有多少人有获胜的可能 分析:只需要排一下序,判断$a[i]-a[i+1]>m$的个数即可.因为排序后,相邻的如果都无法满足这个条件,那他一定被淘汰:反之,他就*有可能*获胜. #include <bits/stdc++.h> using namespace std; typ…

[Link]: [Description] [Solution] 把所有人的能力从大到小排; 能力最大的肯定可能拿冠军; 然后一个一个地往后扫描; 一旦出现a[i-1]-a[i]>k; 则说明从这以后的人,都不可能再和有实力拿冠军的人竞争了 无论怎么安排都赢不了那部分可能拿冠军的人. [NumberOf WA] [Reviw] [Code] #include <bits/stdc++.h> using namespace std; #define lson l,m,rt<<1…