B. String time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output One day in the IT lesson Anna and Maria learned about the lexicographic order. String x is lexicographically less than string y, i…

题目传送门 /* 图论/暴力:这是个连通的问题,每一次把所有度数为1的砍掉,把连接的点再砍掉,总之很神奇,不懂:) */ #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; ; const int INF = 0x3f3f3f3f; int cnt[MAXN]; int a[MAXN]; bool g[MAXN][MAX…

题目传送门 /* BFS:三维BFS,坐标再加上步数,能走一个点当这个地方在步数内不能落到.因为雕像最多8步就会全部下落, 只要撑过这个时间就能win,否则lose */ #include <cstdio> #include <algorithm> #include <queue> #include <vector> #include <cstring> using namespace std; ; const int INF = 0x3f3f3…

A - Chord 题意:就是环中有12个字符,给你三个字符,判断他们之间的间隔,如果第一个和第二个间隔是3并且第二个和第三个间隔是4,那么就输出minor,如果第一个和第二个间隔是4并且第二个和第三个间隔是3,那么就输出major,否则输出strange 思路:用一个数组a中相对位置存入1,因为1个循环是12,让你判断的总长度是7,所以不会产生干扰,直接进行a[i].a[i+3].a[i+7]判断即可 代码: 1 #include<iostream> 2 #include<algori…

题意:给你一个字符串,找第k大的子字符串.(考虑相同的字符串) 题解:建sam,先预处理出每个节点的出现次数,然后处理出每个节点下面的出现次数,然后在dfs时判断一下往哪边走即可,注意一下num会爆int //#pragma GCC optimize(2) //#pragma GCC optimize(3) //#pragma GCC optimize(4) //#pragma GCC optimize("unroll-loops") //#pragma comment(linker,…

D. Numbers time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output One day Anna got the following task at school: to arrange several numbers in a circle so that any two neighboring numbers differs…

C. Statues time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output In this task Anna and Maria play a game with a very unpleasant rival. Anna and Maria are in the opposite squares of a chessboard…

B. Students and Shoelaces time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output Anna and Maria are in charge of the math club for junior students. When the club gathers together, the students be…

B.Phone Numbers 思路:就是简单的结构体排序,只是这里有一个技巧,就是结构体存储的时候,直接存各种类型的电话的数量是多少就行,在读入电话的时候,既然号码是一定的,那么就直接按照格式%c读取就好,在比较的时候也容易比较,直接比较ASCII码就行 代码: #include<iostream> #include<algorithm> #include<cstring> #include<cmath> #include<cstdio> us…

Codeforces Beta Round #76 (Div. 2 Only) http://codeforces.com/contest/94 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb emplac…

Codeforces Beta Round #72 (Div. 2 Only) http://codeforces.com/contest/84 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb emplac…

Codeforces Beta Round #40 (Div. 2) http://codeforces.com/contest/41 A #include<bits/stdc++.h> using namespace std; typedef long long ll; #define maxn 1000006 int main(){ #ifndef ONLINE_JUDGE // freopen("input.txt","r",stdin); #en…

Codeforces Beta Round #80 (Div. 2 Only) A Blackjack1 题意 一共52张扑克,A代表1或者11,2-10表示自己的数字,其他都表示10 现在你已经有一个queen了,问你有多少种方案,能够得到n点 题解 其实只用考虑三种情况,都考虑一下就好了,实在不行就暴力枚举-- 代码 #include<bits/stdc++.h> using namespace std; int n; int main() { scanf("%d",&…

Codeforces Beta Round #62 A Irrational problem 题意 f(x) = x mod p1 mod p2 mod p3 mod p4 问你[a,b]中有多少个数满足f(x)=x 题解 显然直接带进去就好了-- 代码 #include<bits/stdc++.h> using namespace std; int main() { int p1,p2,p3,p4,a,b; scanf("%d%d%d%d%d%d",&p1,&am…

Codeforces Beta Round #83 (Div. 1 Only) A. Dorm Water Supply 题意 给你一个n点m边的图,保证每个点的入度和出度最多为1 如果这个点入度为0,那么这个点就是水龙头点. 如果这个点的出度为0,那么这个点就是储存点. 现在让你把所有水龙头到储存点的路径都输出出来,且输出这条路径的边权最小值 题解 显然是个仙人掌图,所以直接XJB暴力就好了 代码 #include<bits/stdc++.h> using namespace std; co…

题目大意 给一个数列,长度不超过 5000,每次可以将其中的一个数加 1 或者减 1,问,最少需要多少次操作,才能使得这个数列单调不降 数列中每个数为 -109-109 中的一个数 做法分析 先这样考虑:如果操作的次数最少,那么最终得到的不降的数列,必然是由原始数列中的数组成的,具体的证明可以使用反证法 知道了上面讲述的性质,这题就好搞了 先将原始数列(设为 A,共 n 个数)中所有的数去重并从小到达排序,保存在另一个数列中(设为 B,共 m 个数) 定义状态:f[i][j] 表示将原始数列中的…

Codeforces Beta Round #79 (Div. 2 Only) http://codeforces.com/contest/102 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb empla…

Codeforces Beta Round #77 (Div. 2 Only) http://codeforces.com/contest/96 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb emplac…

Codeforces Beta Round #75 (Div. 2 Only) http://codeforces.com/contest/92 A #include<iostream> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define maxn 100005 typedef long long ll; t…

Codeforces Beta Round #74 (Div. 2 Only) http://codeforces.com/contest/90 A #include<iostream> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define maxn 100005 typedef long long ll; t…

Codeforces Beta Round #73 (Div. 2 Only) http://codeforces.com/contest/88 A 模拟 #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb emp…

Codeforces Beta Round #70 (Div. 2) http://codeforces.com/contest/78 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb emplace_bac…

Codeforces Beta Round #69 (Div. 2 Only) http://codeforces.com/contest/80 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb emplac…

Codeforces Beta Round #67 (Div. 2) http://codeforces.com/contest/75 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb emplace_bac…

Codeforces Beta Round #65 (Div. 2) http://codeforces.com/contest/71 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb emplace_bac…

Codeforces Beta Round #63 (Div. 2) http://codeforces.com/contest/69 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb emplace_bac…

Codeforces Beta Round #61 (Div. 2) http://codeforces.com/contest/66 A 输入用long double #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define…

Codeforces Beta Round #59 (Div. 2) http://codeforces.com/contest/63 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb emplace_bac…

Codeforces Beta Round #57 (Div. 2) http://codeforces.com/contest/61 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb emplace_bac…

Codeforces Beta Round #55 (Div. 2) http://codeforces.com/contest/59 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define sqr(x) ((x)*(x)) #define pb push_back #define eb emplace_bac…