首先考虑如何判断当头部位于\(k_0\)时是否可以抓住所有\(n\)个宝物。显然可以排序后贪心将触手与宝藏配对。然后考虑怎样的\(k_0\)作为分界点，即头部位于\(k_0\)满足条件而头部位于\(k_0+1\)不满足条件和头部位于\(k_0\)不满足条件而头部位于\(k_0+1\)满足条件的所

目录2023百度之星初赛三-1石碑文[ABC318-]2023百度之星初赛三-1石碑文[ABC318-]

AtCoder-ABC318AFullMoon暴力枚举判断。提交记录：Submission-AtCoderAtCoder-ABC318BOverlappingSheets暴力枚举判断。提交记录：Submission-AtCoderAtCoder-ABC318CBlueSpring使用通票一定是用在最大的\(kd\)天，排序后枚举\(k\)即可。提交记录：Submission-AtC

#include<bits/stdc++.h>usingnamespacestd;#defineintlonglong#defineendl'\n'intn,a[300010],c[300010],t[300010],s;signedmain(){ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);cin>>n;for(intx,i=1;i<=n;i++){

A-FullMoonhttps://atcoder.jp/contests/abc318/tasks/abc318_aProblemStatementTakahashilikesfullmoons.Lettodaybeday1.ThefirstdayonoraftertodayonwhichhecanseeafullmoonisdayM.Afterthat,hecanseeafullmooneveryPdays,

AT_abc318_eSandwiches题解Links洛谷AtCoderDescription给定一个长度为\(n\)的序列\(a\)，找到满足以下条件的三元组\((a,b,c)\)的数量。\(i<j<k\)；\(a_{i}=a_{k}\)；\(a_{i}\neqa_{j}\)。数据范围：\(1\leqn\leq3\times10^{5}\)，\(1\leqa_{i}\leqn

T1：FullMoon模拟代码实现n,m,p=map(int,input().split())ans=0i=mwhilei<=n:ans+=1i+=pprint(ans)或者答案是\(\lfloor\frac{n+(p-m)}{p}\rfloor\)T2：Overlappingsheets模拟代码实现#include<bits/stdc++.h>#definerep(i,n)f