题目来源:http://poj.org/problem?id=1013 题目大意:有12枚硬币,其中有一枚假币.所有钱币的外表都一样,所有真币的重量都一样,假币的重量与真币不同,但我们不知道假币的重量是比真币轻还是重.现在有一个很准确的天平,我们可以用这个天平称3次来找到那枚假币.只要仔细选择三次称的方式,总可以再三次之内找出那枚假币. 输入:第一行一个正整数n表示样例个数.接下来每三行为一个测试样例.每行为一次称的结果.每枚硬币被编号为A--L.称量的结果有三种,分别用“up”.“down”和…

Counterfeit Dollar Time Limit: 1 Sec  Memory Limit: 64 MB Submit: 415  Solved: 237 Description Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit even though its color a…

 Counterfeit Dollar Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u Submit Status Description Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counter…

Counterfeit Dollar Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 36206   Accepted: 11561 Description Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit ev…

Counterfeit Dollar Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 41559   Accepted: 13237 Description Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit ev…

1.链接地址: http://poj.org/problem?id=1013 http://bailian.openjudge.cn/practice/2692 http://bailian.openjudge.cn/practice/1013 2.题目: Counterfeit Dollar Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 37454   Accepted: 11980 Description Sally…

Counterfeit Dollar Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 42028   Accepted: 13369 Description Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit ev…

http://poj.org/problem?id=1013 (题目链接) 题意 12个硬币中有1个是假的,给出3次称重结果,判断哪个硬币是假币,并且判断假币是比真币中还是比真币轻. Solution 很久以前写的题了,现在翻了翻发现思路还是不错的. http://blog.csdn.net/lyy289065406/article/details/6661421 细节 像这种比较水的与字符串相关的题目用string做一些处理会方便很多,然而这道题好像都差不多. 代码 // poj1013 #i…

Description Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit even though its color and size make it indistinguishable from the real silver dollars. The counterfeit coi…

Description Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit even though its color and size make it indistinguishable from the real silver dollars. The counterfeit coi…

#include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <iostream> #include <string> #include <algorithm> using namespace std; #define FOR(i,s,t) for (int i = s; i < t; ++i) #define ME…

题目链接:http://poj.org/problem?id=1013 解题报告: 1.由于次品的重量不清楚,用time['L'+1]来记录各个字母被怀疑的次数.为负数则轻,为正数则重. 2.用zero['L'+1]记录当天平结果是even时,硬币绝对是真,true; #include <iostream> #include <cmath> using namespace std; int main() { int t; cin>>t; while(t--) { ][]…

时间限制:1 秒 内存限制:32 兆 特殊判题:否 提交:485 解决:215 题目描述: Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit even though its color and size make it indistinguishable from the real s…

http://poj.org/problem?id=1013 #include<stdio.h> #include<string.h> #include<math.h> <<; int main() { ][],lw[][],vis[][]; ],t; scanf("%d",&t); while(t--) { memset(ans,,sizeof(ans)); ; i < ; i ++) { scanf("%s %…

Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit even though its color and size make it indistinguishable from the real silver dollars. The counterfeit coin has a diff…

做了一天水题,挑几个还算凑合的发上来. POJ1008 Maya Calendar 分析: #include <iostream> #include <cstdio> #include <cstring> #include <cmath> using namespace std; ] = {"pop", "no", "zip", "zotz", "tzec",…

在算法竞赛中,很多问题是来不及用数学公式推导出来的.或者说根本就找不到数学规律,这时我们就需要使用枚举来暴力破解. 不过枚举也是需要脑子的,一味的暴力只能超时.因此我这里选择了几道mooc上经典的题目来做复习. 1.完美立方. 思路: 从2到N枚举a的值,2到a枚举d的值,2到d的枚举b的值,2到c枚举b的值.当 满足a*a*a==b*b*b+c*c*c+d*d*d 的时候对结果输出. 代码如下: #include <iostream> using namespace std; int mai…

Counterfeit Dollar Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 50515   Accepted: 15808 Description Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit ev…

Counterfeit Dollar Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 52474   Accepted: 16402 Description Sally Jones has a dozen Voyageur silver dollars. However, only eleven of the coins are true silver dollars; one coin is counterfeit ev…

POJ 排序的思想就是根据选取范围的题目的totalSubmittedNumber和totalAcceptedNumber计算一个avgAcceptRate. 每一道题都有一个value,value = acceptedNumber / avgAcceptRate + submittedNumber. 这里用到avgAcceptedRate的原因是考虑到通过的数量站的权重可能比提交的数量占更大的权重,所以给acceptedNumber乘上了一个因子. 当然计算value还有别的方法,比如POJ上…

Log 2016-3-21 网上找的POJ分类,来源已经不清楚了.百度能百度到一大把.贴一份在博客上,鞭策自己刷题,不能偷懒!! 初期: 一.基本算法: (1)枚举. (poj1753,poj2965) (2)贪心(poj1328,poj2109,poj2586) (3)递归和分治法. (4)递推. (5)构造法.(poj3295) (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996) 二.图算法: (1)图的深度优先遍历和广度优先遍历. (2)最短路…

HDU 模拟题, 枚举1002 1004 1013 1015 1017 1020 1022 1029 1031 1033 1034 1035 1036 1037 1039 1042 1047 1048 1049 1050 1057 1062 1063 1064 1070 1073 1075 1082 1083 1084 1088 1106 1107 1113 1117 1119 1128 1129 1144 1148 1157 1161 1170 1172 1177 1197 1200 1201…

初期:一.基本算法:     (1)枚举. (poj1753,poj2965)     (2)贪心(poj1328,poj2109,poj2586)     (3)递归和分治法.     (4)递推.     (5)构造法.(poj3295)     (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996)二.图算法:     (1)图的深度优先遍历和广度优先遍历.     (2)最短路径算法(dijkstra,bellman-ford,floyd,hea…

初期: 一.基本算法:      (1)枚举. (poj1753,poj2965)      (2)贪心(poj1328,poj2109,poj2586)      (3)递归和分治法.      (4)递推.      (5)构造法.(poj3295)      (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996) 二.图算法:      (1)图的深度优先遍历和广度优先遍历.      (2)最短路径算法(dijkstra,bellman-ford…

leetcode代码 利用堆栈:http://oj.leetcode.com/problems/evaluate-reverse-polish-notation/http://oj.leetcode.com/problems/longest-valid-parentheses/ (也可以用一维数组,贪心)http://oj.leetcode.com/problems/valid-parentheses/http://oj.leetcode.com/problems/largest-rectang…

poj 题目分类 按照ac的代码长度分类(主要参考最短代码和自己写的代码) 短代码:0.01K--0.50K:中短代码:0.51K--1.00K:中等代码量:1.01K--2.00K:长代码:2.01K以上. 短:1147.1163.1922.2211.2215.2229.2232.2234.2242.2245.2262.2301.2309.2313.2334.2346.2348.2350.2352.2381.2405.2406: 中短:1014.1281.1618.1928.1961.2054…

杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze 广度搜索1006 Redraiment猜想 数论:容斥定理1007 童年生活二三事 递推题1008 University 简单hash1009 目标柏林 简单模拟题1010 Rails 模拟题(堆栈)1011 Box of Bricks 简单题1012 IMMEDIATE DECODABILITY…

本文来自:http://www.cppblog.com/snowshine09/archive/2011/08/02/152272.spx 多版本的POJ分类 流传最广的一种分类: 初期: 一.基本算法: (1)枚举. (poj1753,poj2965) (2)贪心(poj1328,poj2109,poj2586) (3)递归和分治法. (4)递推. (5)构造法.(poj3295) (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996) 二.图算法:…

在百度文库上找到的,不知是哪位大牛整理的,真的很不错! zz题 目分类 Posted by fishhead at 2007-01-13 12:44:58.0 -------------------------------------------------------------------------------- acm.pku.edu.cn 1. 排序 1423, 1694, 1723, 1727, 1763, 1788, 1828, 1838, 1840, 2201, 2376, 23…

初期:一.基本算法:     (1)枚举. (poj1753,poj2965)     (2)贪心(poj1328,poj2109,poj2586)     (3)递归和分治法.     (4)递推.     (5)构造法.(poj3295)     (6)模拟法.(poj1068,poj2632,poj1573,poj2993,poj2996)二.图算法:     (1)图的深度优先遍历和广度优先遍历.     (2)最短路径算法(dijkstra,bellman-ford,floyd,hea…