D. Flowers time limit per test 1.5 seconds memory limit per test 256 megabytes input standard input output standard output We saw the little game Marmot made for Mole's lunch. Now it's Marmot's dinner time and, as we all know, Marmot eats flowers. At…
题目链接:http://codeforces.com/problemset/problem/474/D 用RW组成字符串,要求w的个数要k个连续出现,R任意,问字符串长度为[a, b]时,字符串的种类有多少. 递推,dp[i]表示长度为i的种类有多少.当i < k的时候 dp[i] = 1 , 当i == k的时候 dp[i] = 2 ,  否则 dp[i] = dp[i - 1] + dp[i - k] . #include <bits/stdc++.h> using namespac…
We saw the little game Marmot made for Mole's lunch. Now it's Marmot's dinner time and, as we all know, Marmot eats flowers. At every dinner he eats some red and white flowers. Therefore a dinner can be represented as a sequence of several flowers, s…
D. Flowers   We saw the little game Marmot made for Mole's lunch. Now it's Marmot's dinner time and, as we all know, Marmot eats flowers. At every dinner he eats some red and white flowers. Therefore a dinner can be represented as a sequence of sever…
#include<bits/stdc++.h>using namespace std;const long long mod=998244353;int n;int a[100007];long long dp[100007][207][3];//第i位值为j时k是否成立,k=0,i<i-1,k=1,i==i-1,k=2,i>i-1 int main(){    scanf("%d",&n);    for(int i=1;i<=n;i++)   …
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"…
题目链接:http://codeforces.com/contest/283/problem/B 思路: dp[now][flag]表示现在在位置now,flag表示是接下来要做的步骤,然后根据题意记忆化搜索记忆,vis数组标记那些已经访问过的状态. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define REP(i, a, b) for (i…
题目地址:http://codeforces.com/contest/474 A题:Keyboard 模拟水题. 代码例如以下: #include <iostream> #include <cstdio> #include <string> #include <cstring> #include <stdlib.h> #include <math.h> #include <ctype.h> #include <que…
A. Keyboard 题意:一个人打字,可能会左偏一位,可能会右偏一位,给出一串字符,求它本来的串 和紫书的破损的键盘一样 #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<algorithm> using namespace std; typedef long long LL; ]="qwertyuiopasdfghjkl;…
题目链接:http://codeforces.com/contest/474/problem/F 题意简而言之就是问你区间l到r之间有多少个数能整除区间内除了这个数的其他的数,然后区间长度减去数的个数就是答案. 要是符合条件的话,那这个数的大小一定是等于gcd(a[l]...a[r]). 我们求区间gcd的话,既可以利用线段树性质区间递归下去然后返回求解,但是每次查询是log的,所以还可以用RMQ,查询就变成O(1)了. 然后求解区间内有多少个数的大小等于gcd的话,也是利用线段树的性质,区间递…