1057 - Collecting Gold】的更多相关文章

1057 - Collecting Gold   PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB Finally you found the city of Gold. As you are fond of gold, you start collecting them. But there are so much gold that you are getting tired collecti…
题目链接:http://www.lightoj.com/volume_showproblem.php?problem=1057 题解:看似有点下记忆话搜索但是由于他是能走8个方向的也就是说两点的距离其实就是最大的x轴或y轴的差.然后只有15个藏金点状压一下加dfs就行了. #include <iostream> #include <cstring> #include <cstdio> #include <cmath> #define inf 0X3f3f3f…
题目大意: 给你一个矩阵,'x'是你的起始位置, 'g'是宝藏的位置,问最少多少步可以把所有的宝藏取完,并且最后返回起始位置. 注意:没有宝藏的时候输出 0   ==================================================================================== #include<cstdio> #include<cstring> #include<iostream> #include<algor…
题目链接:http://lightoj.com/volume_showproblem.php?problem=1057 题目大意:在二维矩阵中,给你一个起点和至多15个的目标点.要你求出从起点出发经过完所有的点后回到起点的最短路径值.每个点一步可以向 八个方向走. 算法思路:一看就觉得是tsp,用状态压缩.而任意两点的距离就是相应横纵坐标差的较大值.具体看代码. 代码: #include<cstdio> #include<cstring> #include<iostream&…
这道题可以想到几点: 整个行程可以看作一次次的行走,每次行走都是用最短的路程从某一非空点到达另外一非空点: 两点间最少的步数是二者x和y坐标差的最大值: 返回原点这个过程,肯定是取完最后一个黄金后直接用最少的步数从这儿出发回到原点. 然后就是状压DP了: dp[u][S]:经过非空点集S后到达u点最少的步数 转移就枚举从哪儿到达u点的. #include<cstdio> #include<cstring> #include<algorithm> using namesp…
原题链接在这里:https://leetcode.com/problems/path-with-maximum-gold/ 题目: In a gold mine grid of size m * n, each cell in this mine has an integer representing the amount of gold in that cell, 0 if it is empty. Return the maximum amount of gold you can colle…
题目如下: In a gold mine grid of size m * n, each cell in this mine has an integer representing the amount of gold in that cell, 0 if it is empty. Return the maximum amount of gold you can collect under the conditions: Every time you are located in a cel…
Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 5090   Accepted: 2529 Case Time Limit: 2000MS   Special Judge Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material stuff, he collect…
3039: 玉蟾宫 Time Limit: 2 Sec  Memory Limit: 128 MBSubmit: 753  Solved: 444[Submit][Status][Discuss] Description 有一天,小猫rainbow和freda来到了湘西张家界的天门山玉蟾宫,玉蟾宫宫主蓝兔盛情地款待了它们,并赐予它们一片土地.这片土地被分成N*M个格子,每个格子里写着'R'或者'F',R代表这块土地被赐予了rainbow,F代表这块土地被赐予了freda.现在freda要在这里卖…
题目描述 Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material stuff, he collects software bugs. When Ivan gets a new program, he classifies all possible bugs into n categories. Each day he discovers exactly one…
1057: 输入两个整数,求他们相除的余数 Time Limit: 1 Sec  Memory Limit: 128 MBSubmit: 177  Solved: 136[Submit][Status][Web Board] Description 输入两个整数,求他们相除的余数.用带参的宏或模板函数来实现,编程序. Input 两个短整型数a,b 两个长整型数c,d Output a/b的余数 c/d的余数 Sample Input 10 9 100000000 999999 Sample O…
http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1119 1119: Collecting Coins Time Limit: 3 Sec  Memory Limit: 128 MBSubmit: 144  Solved: 35[Submit][Status][Web Board] Description In a maze of r rows and c columns, your task is to collect as many coin…
安装的插件:Activiti 在Eclipse安装插件时,报以下错误: An error occurred while collecting items to be installed session context was:(profile=epp.package.jee, phase=org.eclipse.equinox.internal.p2.engine.phases.Collect, operand=, action=). Problems downloading artifact:…
Problem Description You are given N baskets of gold coins. The baskets are numbered from 1 to N. In all except one of the baskets, each gold coin weighs w grams. In the one exceptional basket, each gold coin weighs w-d grams. A wizard appears on the…
Problem Introduction This problem is about implementing an algorithm for the knapsack without repetitions problem. Problem Description Task.In this problem, you are given a set of bars of gold and your goal is to take as much gold as possible into yo…
Decsription 给你一个矩阵,求最大了 01相间 的矩阵. Sol DP+悬线法. 这是一个论文啊 <浅谈用极大化思想解决最大子矩形问题>--王知昆. 枚举每一根悬线,记录最左/右/上能到达的点,统计答案. Code /************************************************************** Problem: 1057 User: BeiYu Language: C++ Result: Accepted Time:1384 ms Me…
[POJ2096]Collecting Bugs Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material stuff, he collects software bugs. When Ivan gets a new program, he classifies all possible bugs into n categories. E…
题目链接:51nod 1057 N的阶乘 #include<cstdio> using namespace std; typedef long long ll; ; const int mod = 1e8; ll a[N] = {}; int n; int main(){ int i, j, c, cnt; scanf("%d", &n); cnt = ; ; j <= n; ++j){ ; i <cnt; ++i){ a[i] = a[i] * j +…
题目描述 Orz Panda is addicted to one RPG game. To make his character stronger, he have to fulfil tasks to get EXP for higher level.At first he accepted all the tasks.But after he read rules carefully, he realized that what he has done was stupid. Rule:…
Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 1716   Accepted: 783 Case Time Limit: 2000MS   Special Judge Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material st…
Gold Balanced Lineup Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 10924 Accepted: 3244 Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by h…
1057 N的阶乘 基准时间限制:1 秒 空间限制:131072 KB 输入N求N的阶乘的准确值. Input 输入N(1 <= N <= 10000) Output 输出N的阶乘 Input示例 5 Output示例 120 import java.math.BigInteger; import java.util.Scanner; public class Main { static BigInteger ans(BigInteger a,int b){ BigInteger anss=B…
Gold Balanced Lineup Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 13215 Accepted: 3873 Description Farmer John's N cows (1 ≤ N ≤ 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by h…
Gold Coins Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 21767   Accepted: 13641 Description The king pays his loyal knight in gold coins. On the first day of his service, the knight receives one gold coin. On each of the next two days…
Gold miner Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1889    Accepted Submission(s): 740Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice …
Collecting Bugs Time Limit: 10000MS   Memory Limit: 64000K Total Submissions: 3064   Accepted: 1505 Case Time Limit: 2000MS   Special Judge Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material s…
问题描述: Pots of gold game: Two players A & B. There are pots of gold arranged in a line, each containing some gold coins (the players can see how many coins are there in each gold pot - perfect information). They get alternating turns in which the play…
数位DP 2009年刘聪<浅谈数位类统计问题> 例题一 从组合数 以及 数位DP的角度都可以做…… 首先转化成求1~n内K进制下只有0.1的数的个数: 考虑K进制下第一个为1的位,剩下的数位中0和1随便放……也就是算组合数了. 然后区间相减…… 写的好乱……还是去看论文吧 //Ural 1057 #include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #inc…
http://www.bnuoj.com/bnuoj/problem_show.php?pid=1057 [题意]:给定x的值,带入f(x)求函数值 [题解]:注意第一个数的符号可能是'+',这里把我坑死了... [code]: #include <iostream> #include <stdio.h> #include <math.h> #include <string.h> #include <algorithm> using namesp…
Description Ivan is fond of collecting. Unlike other people who collect post stamps, coins or other material stuff, he collects software bugs. When Ivan gets a new program, he classifies all possible bugs into n categories. Each day he discovers exac…