· The 19th Heilongjiang Provincial Collegiate Programming Contest · The 10th Jimei University Programming Contest · 2022 Shanghai Collegiate Programming Contest · 2022 International Collegiate Programming
代码: #include<bits/stdc++.h>usingnamespacestd;usingLL =longlong;intmain(){ ios::sync_with_stdio(false);cin.tie(0); string E, C; cin >> E >> C; map<char,int> B, P;for(inti =0; i <8; i ++ ){if(C[i] =='B') B[E[i]] ++ ;if(C[i] =='P') P[E[i]] ++ ...
vp的时候搞了个枚举+二分卡了过去。其实是个类似lcs的dp。dp[i][j]表示s到i,F匹配到第j位的最短长度。 若s[i] != f[j], 则dp[i][j] = dp[i - 1][j] + 1 若s[i] == f[j],则dp[i][j] = min(dp[i - 1][j], dp[i - 1][j - 1] + 1) 代码: constintN=100005;int...
2022 icpc 济南 (2022 International Collegiate Programming Contest, Jinan Site) 链接:https://codeforces.com/gym/104076 A. Tower 枚举最后的取值,然后计算每个数变成这个取值的最⼩次数,去掉最大的m个,取min。 #include"bits/stdc++.h"usingnamespacestd;usingi64=longlong;voidsolve(){intn,m;cin>>n>...
[Shanghai, China, August 22, 2022] The 2022 ICPC (International Collegiate Programming Contest) Shanghai Training Camp powered by Huawei ended on August 21, drawing praise from participants and involved universities. The intense, three-day competition and training program had nearly 100 part...
The Universities of Alberta, Calgary, and Lethbridge would like to invite you to participate in the open division of theAlberta Collegiate Programming Contest 2022. The contest starts onNovember 26 at 11:30am mountain time. The contest link will be atacpc22open.kattis.com. It currently returns...
2022ccpc绵阳站 2022 China Collegiate Programming Contest (CCPC) Mianyang Onsite,C.CatchYouCatchMe题目大意:给你n条路径构成一个无向树,结点编号为1~n。在这棵树中,结点1为出口,其他所有结点上都有一只蝴蝶。每一分钟,每只蝴蝶都会沿着一条树的
晚上花了3个小时打了一把湖北省赛,感觉被东北省赛出的要好的多的多的多了,至少都是在贴合算法而不是思维,虽然比较难但是体验也都还好,最后是6题稳在6题党第二。 进入正题: A. Nucleic Acid Test 图论大杂烩 题意: 给定一张连通的带权无向图,有n个建筑,其中有k个可以做核算的医院。我们的目标是走遍所...
【ZJCPC2022 第19届 浙江省赛】The 19th Zhejiang Provincial Collegiate Programming Contest(CBALGMIF 8题),首先通过Hash或马拉车等方式O(1)特判起始串为回文串的情况。+对于接
说是阅读理解题是因为只要理解好题意你就会发现什么什么概率都是诈骗。 CODE voidsolve(){intn =0, m =0; std::cin >> n >> m;for(inti =0; i < m; i++) { cnt[i] =1; } std::string s;for(inti =0; i < n; i++) {