题目链接:http://codeforces.com/problemset/problem/460/C C. Present time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little beaver is a beginner programmer, so informatics is his favorite subjec…

Codeforces Round #262 (Div. 2) 1003 C. Present time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little beaver is a beginner programmer, so informatics is his favorite subject. Soon his info…

Codeforces Round #262 (Div. 2) 1004 D. Little Victor and Set time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Victor adores the sets theory. Let us remind you that a set is a group of…

题目链接 B Little Dima and Equation 题意:给a, b,c 给一个公式,s(x)为x的各个位上的数字和,求有多少个x. 分析:直接枚举x肯定超时,会发现s(x)范围只有只有1-81,所以枚举一下就行. 在做题的时候,用了pow()错了3次,反正以后不用pow了,还是手写吧.会有误差.pow返回的是double型的. 昨天在b题耽误了好多时间,先是提交错第一组,然后又被人cha了.注意在x在1-10^9之间. #include <iostream> #include &…

这是最大化最小值的一类问题,这类问题通常用二分法枚举答案就行了. 二分答案时,先确定答案肯定在哪个区间内.然后二分判断,关键在于怎么判断每次枚举的这个答案行不行. 我是用a[i]数组表示初始时花的高度,b[i]表示要达到当前枚举的答案(即mid的值)需要这朵花再涨多少.这两个数组很好算,关键是一次浇连续的w朵花,如何更新区间(暴力的O(n2)的去更新就超时了)?可以用线段树,但是这道题没有涉及区间查询,就是在一个数组上更新区间,用线段树未免小题大做.那么其实这种更新就用延迟标记的思想(懒操作)就…

题目链接 A. Vasya and Socks time limit per test:2 secondsmemory limit per test:256 megabytesinput:standard inputoutput:standard output Vasya has n pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. Wh…

https://codeforces.com/contest/1121/problem/F 题意 给你一个有n(<=5000)个字符的串,有两种压缩字符的方法: 1. 压缩单一字符,代价为a 2. 压缩一个串,条件是这个串是前面整个串的连续子串,代价为b 题解 n<=5000 定义dp[i]为压缩前i个字符的代价,则答案为dp[n] dp[i]=min(dp[i-1]+a,min(dp[j]+b)(即[j+1,i]为[1,j]的子串)) 用字符串哈希处理判定一个串是否为前面的子串 坑点 串ab…

题目: C. Present time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Little beaver is a beginner programmer, so informatics is his favorite subject. Soon his informatics teacher is going to have…

详见:http://robotcator.logdown.com/posts/221514-codeforces-round-262-div-2 1:A. Vasya and Socks   http://codeforces.com/contest/460/problem/A 有n双袜子,每天穿一双然后扔掉.每隔m天买一双新袜子,问最多少天后没有袜子穿. . 简单思维题:曾经不注重这方面的训练,结果做了比較久.这样的题自己边模拟边想.只是要多考虑trick ```c++ int main(){…

A. Vasya and Socks time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasya has n pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When…

D. Exams Problem Description: Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m. About every day we know exam for which one of m subjects can be passed on that day. Perhap…

A #include <iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<stdlib.h> #include<vector> #include<cmath> #include<queue> #include<set> using namespace std; #define N 100000 #def…

D. Exams time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasiliy has an exam period which will continue for n days. He has to pass exams on m subjects. Subjects are numbered from 1 to m. Ab…

题目链接:http://www.codeforces.com/problemset/problem/460/A题意:Vasya每天用掉一双袜子,她妈妈每m天给他送一双袜子,Vasya一开始有n双袜子,请问第几天的时候Vasya会没有袜子穿?C++代码: #include <iostream> using namespace std; int n, m; int main() { cin >> n >> m; ; while (n) { d ++; if (n) n --…

E. Roland and Rose Time Limit: 1 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/problemset/problem/460/E Description Roland loves growing flowers. He has recently grown a beautiful rose at point (0, 0) of the Cartesian coordinate system. The ro…

#include<bits/stdc++.h>using namespace std;const long long N=1e5+5;const long long MOD=1e9+7;long long n,x,y,ans=0;long long cost[N];pair<long long,long long>a[N];multiset<pair<pair<long long,long long>,long long> >s;int main…

题目链接:http://codeforces.com/contest/460/problem/A A. Vasya and Socks time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasya has n pairs of socks. In the morning of each day Vasya has to put o…

题目: B. Little Dima and Equation time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Little Dima misbehaved during a math lesson a lot and the nasty teacher Mr. Pickles gave him the following pr…

A. Vasya and Socks time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vasya has n pairs of socks. In the morning of each day Vasya has to put on a pair of socks before he goes to school. When…

题意: 求n个数中两两和的异或. 思路: 逐位考虑,第k位只需考虑0~k-1位,可通过&(2k+1-1)得到一组新数. 将新数排序,当两数和在[2k,2k+1)和[2k+1+2k,2k+2)之间时该位为1,又因为两数的最大和为2*(2k+1-1)=2k+2-2, 即当两数和在[2k,2k+1)和[2k+1+2k,2k+2-2]之间时该位为1. 对于每个数,找到和 大于等于2k 小于2k+1 大于等于2k+1+2k 的三个临界点(因为两数之和一定小于等于2k+2-2,所以第四个临界点可以忽略),…

// Codeforces Round #365 (Div. 2) // C - Chris and Road 二分找切点 // 题意:给你一个凸边行,凸边行有个初始的速度往左走,人有最大速度,可以停下来,竖直走. // 问走到终点的最短时间 // 思路: // 1.贪心来做 // 2.我觉的二分更直观 // 可以抽象成:一条射线与凸边行相交,判断交点.二分找切点 #include <bits/stdc++.h> using namespace std; #define LL long lon…

题目传送门 /* 二分查找/暴力:先埃氏筛选预处理,然后暴力对于每一行每一列的不是素数的二分查找最近的素数,更新最小值 */ #include <cstdio> #include <cstring> #include <algorithm> using namespace std; ; ; const int INF = 0x3f3f3f3f; int a[MAXN][MAXN]; int mn_r[MAXN]; int mn_c[MAXN]; bool is_prim…

Codeforces Round #404 (Div. 2) 题意:对于 n and m (1 ≤ n, m ≤ 10^18)  找到 1) [n<= m] cout<<n; 2) [n>m]最小的 k => (k -m) * (k-m+1) >= (n-m)*2 成立 思路:二分搜索 #include <bits/stdc++.h> #include <map> using namespace std; #define LL long long…

A: 题目大意: 在一个multiset中要求支持3种操作: 1.增加一个数 2.删去一个数 3.给出一个01序列,问multiset中有多少这样的数,把它的十进制表示中的奇数改成1,偶数改成0后和给出的01序列相等(比较时如果长度不等各自用0补齐) 题解: 1.我的做法是用Trie数来存储,先将所有数用0补齐成长度为18位,然后就是Trie的操作了. 2.官方题解中更好的做法是,直接将每个数的十进制表示中的奇数改成1,偶数改成0,比如12345,然后把它看成二进制数10101,还原成十进制是2…

Codeforces Round #271 (Div. 2) A - Keyboard 题意 给你一个字符串,问你这个字符串在键盘的位置往左边挪一位,或者往右边挪一位字符,这个字符串是什么样子 题解 模拟一下就好了 代码 #include<bits/stdc++.h> using namespace std; string s[3]; map<char,int>r,c; char ss[2][107]; int main() { s[0]="qwertyuiop"…

Codeforces Round #113 (Div. 2) B. Polygons 题意 给一个\(N(N \le 10^5)\)个点的凸包 \(M(M \le 2 \cdot 10^4)\)次询问,每次给一个点判断该点是否在凸包内. 思路 按\(y\)坐标将凸包分成两部分. 在左右两边二分找出夹住该点的\(y\)值区间,判断叉积正负. 代码 B. Polygons D. Shoe Store 题意 有\(N \le 10^5\)双鞋,每双鞋价格为\(c_i \le 10^9\),大小为\(s…

Codeforces Round #111 (Div. 2) C. Find Pair 题意 给\(N(N \le 10^5)\)个数,在所有\(N^2\)对数中求第\(K(K \le N^2)\)对数. 排序按照pair比较,first为第一关键字,second第二关键字. 思路 统计\(cnt[x]\)为值\(x\)出现的次数. 第一关键字为\(x\)的对数为\(cnt[x] \times n\),显然可以找到第一关键字. 在确定第一关键字\(x\)后,第二关键字\(y\)的出现次数为\(c…

Codeforces Round #372 (Div. 2) C. Plus and Square Root 题意 一个游戏中,有一个数字\(x\),当前游戏等级为\(k\),有两种操作: '+'按钮:使得\(x=x+k\) '√'按钮:使得\(x=\sqrt{x}\),此时\(x\)必须是平方数,游戏等级加1,即\(k=k+1\),且\(\sqrt{x}\)是\(k+1\)的倍数. 游戏开始时,\(x=2,k=1\),输出\(n(n \le 10^5)\)个数,表示每个等级对应的\(\frac…

Vasiliy's Multiset 题目链接: http://codeforces.com/contest/706/problem/D Description Author has gone out of the stories about Vasiliy, so here is just a formal task description. You are given q queries and a multiset A, initially containing only integer…

Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) 说一点东西: 昨天晚上$9:05$开始太不好了,我在学校学校$9:40$放学我呆到十点然后还要跑回家耽误时间....要不然$D$题就写完了 周末一些成绩好的同学单独在艺术楼上课然后晚上下第一节晚自习和他们在回廊里玩开灯之后再关上一片漆黑真好玩 A.Andryusha and Socks 日常煞笔提.....我竟然$WA$了一次忘了$n<<1$ #include <…