不容易系列之(3)—— LELE的RPG难题 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 38217    Accepted Submission(s): 15342 Problem Description 人称“AC女之杀手”的超级偶像LELE最近忽然玩起了深沉,这可急坏了众多“Cole”(LELE的粉丝,即"可乐"),经…

Problem Description 人称"AC女之杀手"的超级偶像LELE最近忽然玩起了深沉,这可急坏了众多"Cole"(LELE的粉丝,即"可乐"),经过多方打探,某资深Cole终于知道了原因,原来,LELE最近研究起了著名的RPG难题: 有排成一行的n个方格,用红(Red).粉(Pink).绿(Green)三色涂每个格子,每格涂一色,要求任何相邻的方格不能同色,且首尾两格也不同色．求全部的满足要求的涂法. 以上就是著名的RPG难题. 如果…

HDOJ(HDU).2044-2049 递推专题 点我挑战题目 HDU.2044 题意分析 先考虑递推关系:从1到第n个格子的时候由多少种走法? 如图,当n为下方格子的时候,由于只能向右走,所以有2中走法.当n为上方格子的时候,由于只能向右走,所以也有2种走法. 不妨用a[n]来表示第n个格子有几种走法,根据上述描述,不难找出递推关系,a[n] = a[n-1] + a[n-2].但是对于n<2的数字,就不适用了.也很简单,n<2的时候数一下即可,从1走到1,只有一种走法,从1走到2,有一种走…

着色问题,递推,当超过3个块时,规律明显,此时可以是n-2的头尾重复+与头尾不同颜色,也可以是n-1+与头尾均不相同眼色情况.经典递推.注意long long. #include <stdio.h> #define MAXNUM 52 ,,,}; int main() { int n; int i; ; i<MAXNUM; ++i) kinds[i] = kinds[i-]* + kinds[i-]; while (scanf("%d", &n) != EOF…

#include <iostream> using namespace std; ]; int main() { int n; a[] = ; a[] = ; a[] = ; ; i <= ; i++) { a[i] = a[i - ] + * a[i - ]; } while (cin >> n) { cout << a[n] << endl; } }…

FatMouse' Trade Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 56784    Accepted Submission(s): 19009 Problem Description FatMouse prepared M pounds of cat food, ready to trade with the cats g…

Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 3049    Accepted Submission(s): 2364 Problem Description You are the CEO of Nasty Hacks Inc., a company that creates small pieces of…

Box of Bricks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5994    Accepted Submission(s): 2599 Problem Description Little Bob likes playing with his box of bricks. He puts the bricks one up…

Problem Description Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges' favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color…

卧槽....最近刷的cf上有最短路,本来想拿这题复习一下.... 题意就是在输出最短路的情况下,经过每个节点会增加税收,另外要字典序输出,注意a到b和b到a的权值不同 然后就是处理字典序的问题,当松弛时发现相同值的时候,判断两条路径的字典序 代码 #include "stdio.h" const int MAXN=110; const int INF=10000000; bool vis[MAXN]; int pre[MAXN]; int cost[MAXN][MAXN],lowcos…