易得递推式为f[i]=f[i-1]+f[i-M] 最终答案即为f[N]. 由于N很大,用矩阵快速幂求解. code: #include<bits/stdc++.h> using namespace std; typedef long long ll; const int MOD=1e9+7; ll n,m; struct mat { ll a[105][105]; }; mat mat_mul(mat x,mat y) { mat res; memset(res.a,0,sizeof(res.…

Problem   Educational Codeforces Round 60 (Rated for Div. 2) - D. Magic Gems Time Limit: 3000 mSec Problem Description Input The input contains a single line consisting of 2 integers N and M (1≤N≤10^18, 2≤M≤100). Output Print one integer, the total n…

Problem   Educational Codeforces Round 60 (Rated for Div. 2) - C. Magic Ship Time Limit: 2000 mSec Problem Description Input Output The only line should contain the minimal number of days required for the ship to reach the point (x2,y2)(x2,y2). If it…

Educational Codeforces Round 60 (Rated for Div. 2) 题目链接:https://codeforces.com/contest/1117 A. Best Subsegment 题意: 给出n个数,选取一段区间[l,r],满足(al+...+ar)/(r-l+1)最大,这里l<=r,并且满足区间长度尽可能大. 题解: 因为l可以等于r,所以我们可以直接考虑最大值,因为题目要求,直接求连续的最大值的长度就是了. 代码如下: #include <bits…

[Educational Codeforces Round 16]C. Magic Odd Square 试题描述 Find an n × n matrix with different numbers from 1 to n2, so the sum in each row, column and both main diagonals are odd. 输入 The only line contains odd integer n (1 ≤ n ≤ 49). 输出 Print n lines…

A. Best Subsegment 题意 找 连续区间的平均值  满足最大情况下的最长长度 思路:就是看有几个连续的最大值 #include<bits/stdc++.h> using namespace std; ; int a[maxn]; int main(){ int n; scanf("%d",&n); ; ;i<n;i++)scanf("%d",&a[i]),maxnum=max(maxnum,a[i]); ; ; ;i…

题目传送门 题意: 一个魔法水晶可以分裂成m个水晶,求放满n个水晶的方案数(mol1e9+7) 思路: 线性dp,dp[i]=dp[i]+dp[i-m]; 由于n到1e18,所以要用到矩阵快速幂优化 注意初始化 代码: #include<bits/stdc++.h> using namespace std; #define mod 1000000007 typedef long long ll; #define MAX 105 ;//矩阵的大小 int T; ll n,m; ll add(ll…

time limit per test 2 second memory limit per test 256 megabytes input standard inputoutput standard output You a captain of a ship. Initially you are standing in a point (x1,y1)(x1,y1) (obviously, all positions in the sea can be described by cartesi…

F. Magic Matrix 题目连接: http://www.codeforces.com/contest/632/problem/F Description You're given a matrix A of size n × n. Let's call the matrix with nonnegative elements magic if it is symmetric (so aij = aji), aii = 0 and aij ≤ max(aik, ajk) for all…

D. Magic Numbers 题目连接: http://www.codeforces.com/contest/628/problem/D Description Consider the decimal presentation of an integer. Let's call a number d-magic if digit d appears in decimal presentation of the number on even positions and nowhere els…

Magic Numbers 题意:给定长度不超过2000的a,b;问有多少个x(a<=x<=b)使得x的偶数位为d,奇数位不为d;且要是m的倍数,结果mod 1e9+7; 直接数位DP;前两维的大小就是mod m的大小,注意在判断是否f[pos][mod] != -1之前,要判断是否为边界,否则会出现重复计算: #include<bits/stdc++.h> using namespace std; #define inf 0x3f3f3f3f #define MS1(a) mem…

https://codeforces.com/contest/1117/problem/D 题意 有n个特殊宝石(n<=1e18),每个特殊宝石可以分解成m个普通宝石(m<=100),问组成n颗宝石有多少种方法 题解 数据很大:找规律or矩阵快速幂 转移方程: dp[i]=dp[i-1]+dp[i-m] 因为n<=1e18可以用矩阵快速幂 构造矩阵如图: \[ \left[ \begin{matrix} f[i-1] & f[i-2] & \cdots & f[i…

https://codeforces.com/contest/1117/problem/C 题意 在一个二维坐标轴上给你一个起点一个终点(x,y<=1e9),然后给你一串字符串代表每一秒的风向,然后每一秒你也可以选择一个方向移动,问到达从起点到终点的最短时间 题解 思维:等效法 假如到达终点之后,可以静止不动,可以用自己走反向风方向来抵消风的方向 先只考虑风向,用前缀和将每一秒到达的位置维护出来,假设x秒到达的位置是[x,y],终点为[sx,sy],则若|sx-x|+|sy-y|<=x,则一定…

F:考虑对于每个字母对求出删掉哪些字符集会造成字符串不合法,只要考虑相邻出现的该字母对即可,显然这可以在O(np2)(或小常数O(np3))内求出.然后再对每个字符集判断是否能通过一步删除转移而来即可. #include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> using nam…

#include <bits/stdc++.h>using namespace std;const long long mod = 1e9+7;unordered_map<long long,long long>mp;long long n,m;long long dp(long long n){    if(n<0)        return 0;    if(n<m)        return 1;    if(mp.find(n)!=mp.end())//已经…

#include <bits/stdc++.h>using namespace std;int main(){ string t; cin>>t; int n=t.size(); string s1(n,'a'),s2(n,'a'),s3(n,'a'); for(int i=0;i<n;i++){  s1[i]=char('a'+(i%26));//从a到z循环  s2[i]=char('a'+((i/26)%26));//第i位为(i/26)%26+'a',保证了26*26…

题目大意:这是一道交互题.给你一个长度为n的字符串,这个字符串是经过规则变换的,题目不告诉你变换规则,但是允许你提问3次:每次提问你给出一个长度为n的字符串,程序会返回按变换规则变换后的字符串,提问3次后你需要猜出这个字符串.解法是学习https://blog.csdn.net/baiyifeifei/article/details/87807822 这个博主的,借用了进制的思想非常巧妙. 解法:对于某个位置的来源位置我们设为x,因为26*26*26>10000,那么x可以唯一表示为x=a*26…

[Educational Codeforces Round 16]E. Generate a String 试题描述 zscoder wants to generate an input file for some programming competition problem. His input is a string consisting of n letters 'a'. He is too lazy to write a generator so he will manually ge…

[Educational Codeforces Round 16]D. Two Arithmetic Progressions 试题描述 You are given two arithmetic progressions: a1k + b1 and a2l + b2. Find the number of integers x such that L ≤ x ≤ R andx = a1k' + b1 = a2l' + b2, for some integers k', l' ≥ 0. 输入 Th…

[Educational Codeforces Round 16]B. Optimal Point on a Line 试题描述 You are given n points on a line with their coordinates xi. Find the point x so the sum of distances to the given points is minimal. 输入 The first line contains integer n (1 ≤ n ≤ 3·105)…

[Educational Codeforces Round 16]A. King Moves 试题描述 The only king stands on the standard chess board. You are given his position in format "cd", where c is the column from 'a' to 'h' and dis the row from '1' to '8'. Find the number of moves perm…

Educational Codeforces Round 6 C. Pearls in a Row 题意:一个3e5范围的序列:要你分成最多数量的子序列,其中子序列必须是只有两个数相同, 其余的数只能出现一次. 策略: 延伸:这里指的延伸如当发现1-1如果以最后出现重叠的数为右边界则就表示左延伸,若以1.0.1..0第二个0前一个位置作为右边界就为右延伸: 开始时想是右延伸,考虑到可能只出现一组两个数相同,认为向左延伸会出错,但是直接WA了之后,发现这并不是题目的坑点(其实只需将最后一组改成左右…

Educational Codeforces Round 9 Longest Subsequence 题目描述:给出一个序列,从中抽出若干个数,使它们的公倍数小于等于\(m\),问最多能抽出多少个数,并输出方案与公倍数. solution 枚举每一个数,将它的倍数的答案加一,最大值就是答案. 时间复杂度:\(O(nlogn)\) Thief in a Shop 题目描述:给出\(n\)中物品与它们的价值,每种物品都有无限个,问从中拿出\(m\)个,价值有可能是什么,输出所有的价值. soluti…

Educational Codeforces Round 37 这场有点炸,题目比较水,但只做了3题QAQ.还是实力不够啊! 写下题解算了--(写的比较粗糙,细节或者bug可以私聊2333) A. Water The Garden 题意:给你一个长度为\(n\)的池子,告诉你哪些地方一开始有水, 每秒可以向左和向右增加一格的水, 问什么时候全部充满水.(\(n \le 200\)) 题解:按题意模拟.每次进来一个水龙头,就更新所有点的答案 (取\(min\)).最 后把所有点取个\(max\)就…

616A - Comparing Two Long Integers    20171121 直接暴力莽就好了...没什么好说的 #include<stdlib.h> #include<stdio.h> #include<math.h> #include<cstring> #include<iostream> #include<algorithm> using namespace std; string a,b;int sa,sb;…

Educational Codeforces Round 43 (Rated for Div. 2) https://codeforces.com/contest/976 A #include<bits/stdc++.h> using namespace std; #define lson l,mid,rt<<1 #define rson mid+1,r,rt<<1|1 #define IT set<ll>::iterator #define sqr(x)…

Educational Codeforces Round 35 (Rated for Div. 2) https://codeforces.com/contest/911 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 IT set<ll>::iterator #define sqr(…

Codeforces Educational Codeforces Round 44 (Rated for Div. 2) F. Isomorphic Strings 题目连接: http://codeforces.com/contest/985/problem/F Description You are given a string s of length n consisting of lowercase English letters. For two given strings s an…

Codeforces Educational Codeforces Round 44 (Rated for Div. 2) E. Pencils and Boxes 题目连接: http://codeforces.com/contest/985/problem/E Description Mishka received a gift of multicolored pencils for his birthday! Unfortunately he lives in a monochrome w…

Educational Codeforces Round 63 (Rated for Div. 2)题解 题目链接 A. Reverse a Substring 给出一个字符串,现在可以对这个字符串进行一次翻转,问是否存在一种方案,可以使得翻转后字符串的字典序可以变小.   这个很简单,贪心下就行了. 代码如下: Code #include <bits/stdc++.h> using namespace std; typedef long long ll; const int N = 3e5…