https://codeforces.com/problemset/problem/1033/C 一开始觉得自己的答案会TLE,但是吸取徐州赛区的经验去莽了一发. 其实因为下面这个公式是 $O(nlogn)$ 的,不是 $O(n²)$ ,所以这样做是可行的.学到了新的知识. $$\sum\limits_{i=1}^{n}\lfloor\frac{n}{i}\rfloor$$ PS:学会LaTeX啦! #include<bits/stdc++.h> using namespace std; #d…

题目 简单dp //简单的dp #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; ][];//dp[i][j] di i ceng di j ge zui da he ][]; int main() { int t; scanf("%d",&t); while(t--) { int n; scanf("%d",&…

题目链接:http://codeforces.com/problemset/problem/455/A 给你n个数,要是其中取一个大小为x的数,那x+1和x-1都不能取了,问你最后取完最大的和是多少. 简单dp,dp[i]表示取i时zui最大和为多少,方程为dp[i] = max(dp[i - 1] , dp[i - 2] + cont[i]*i). #include <bits/stdc++.h> using namespace std; typedef __int64 LL; ; LL a…

Problem H. ICPC QuestTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100500/attachments Description Noura Boubou is a Syrian volunteer at ACM ACPC (Arab Collegiate Programming Contest) since 2011. She graduated from Tishreen Un…

C. Writing Code Time Limit: 20 Sec  Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/544/problem/C Description Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working o…

题目链接:点击打开链接 给定n*m 的矩阵 常数k 以下一个n*m的矩阵,每一个位置由 0-9的一个整数表示 问: 从最后一行開始向上走到第一行使得路径上的和 % (k+1) == 0 每一个格子仅仅能向↖或↗走一步 求:最大的路径和 最后一行的哪个位置作为起点 从下到上的路径 思路: 简单dp #include <cstdio> #include <algorithm> #include<iostream> #include<string.h> #incl…

D. Fedor and Essay time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After you had helped Fedor to find friends in the «Call of Soldiers 3» game, he stopped studying completely. Today, the E…

C. Dasha and Password time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output After overcoming the stairs Dasha came to classes. She needed to write a password to begin her classes. The password i…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33384    Accepted Submission(s): 15093 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

题目链接 这道题也是简单dp里面的一种经典类型,递推式就是dp[i] = min(dp[i-150], dp[i-200], dp[i-350]) 代码如下: #include<iostream> #include <stdio.h> using namespace std; ]; int main() { ; i < ; i++) dp[i] = i; ; i < ; i++) { int minn; ) dp[i] = dp[i - ]; ) dp[i] = min…