链接: https://codeforces.com/contest/1278/problem/D 题意: As the name of the task implies, you are asked to do some work with segments and trees. Recall that a tree is a connected undirected graph such that there is exactly one simple path between every…

链接: https://codeforces.com/contest/1278/problem/C 题意: Karlsson has recently discovered a huge stock of berry jam jars in the basement of the house. More specifically, there were 2n jars of strawberry and blueberry jam. All the 2n jars are arranged in…

链接: https://codeforces.com/contest/1278/problem/B 题意: You are given two integers a and b. You can perform a sequence of operations: during the first operation you choose one of these numbers and increase it by 1; during the second operation you choos…

链接: https://codeforces.com/contest/1278/problem/A 题意: Polycarp has built his own web service. Being a modern web service it includes login feature. And that always implies password security problems. Polycarp decided to store the hash of the password…

比赛传送门 A. Shuffle Hashing 题意:加密字符串.可以把字符串的字母打乱后再从前面以及后面接上字符串.问加密后的字符串是否符合加密规则. 题解:字符串的长度很短,直接暴力搜索所有情况 // https://codeforces.com/contest/1278/problem/A #include<iostream> #include<cstdio> #include<cstring> using namespace std; int T; char…

题:https://codeforces.com/contest/1278/problem/B 思路:还是把1~n分配给俩个数,让他们最终相等 假设刚开始两个数字相等,然后一个数字向前走了abs(b-a)步,由等差数列求和公式,这时候我们贪心的让另外一个数字走大于等于abs(b - a)的最小步数,然后如果两数相等必须满足走的步数之和%2=0 #include<bits/stdc++.h> using namespace std; #define pb push_back typedef lo…

A题 给出n对串,求s1,是否为s2一段连续子串的重排,串长度只有100,从第一个字符开始枚举,sort之后比较一遍就可以了: char s1[200],s2[200],s3[200]; int main() { int q; cin >> q; while(q--) { cin >> s1 >> s2; int s11 = strlen(s1),s22 = strlen(s2),i; sort(s1,s1+s11); s3[s11] = '\0'; for(i = 0…

链接 直接用数组记录每个字母的个数即可 #include<bits/stdc++.h> using namespace std; int a[26] = {0}; int b[26] = {0}; int judge() { //cout<<"111"<<endl; for (int i = 0; i < 26; ++i) { if (a[i]!=b[i]) { return 0; } } return 1; } int main() { //…

Shuffle Hashing A and B Berry Jam Segment Tree Tests for problem D Cards Shuffle Hashing \[ Time Limit: 2 s\quad Memory Limit: 256 MB \] 处理出 \(s_1\) 中各个字符出现的次数,然后双指针维护 \(s_2\) 中每一段长度为 \(len(s_1)\) 的串中字符出现的次数,如果存在某一段和 \(s_1\) 的字符次数相同,则是答案. view #inclu…

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