http://poj.org/problem?id=3056 The Bavarian Beer Party Time Limit: 6000MS   Memory Limit: 65536K Total Submissions: 995   Accepted: 359 Description The professors of the Bayerische Mathematiker Verein have their annual party in the local Biergarten.…

  http://poj.org/problem?id=1651Multiplication Puzzle   Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6188   Accepted: 3777 Description The multiplication puzzle is played with a row of cards, each containing a single positive integer.…

http://poj.org/problem?id=1679 The Unique MST Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 23339   Accepted: 8284 Description Given a connected undirected graph, tell if its minimum spanning tree is unique. Definition 1 (Spanning Tree…

http://poj.org/problem?id=3278 题目大意: m,n两个数m可+1, -1, *2变成n,需要经过几步 #include<stdio.h> #include<stdlib.h> #include<math.h> #include<string.h> #include<queue> #define max(a, b)(a > b ? a : b) #define N 100010 using namespace s…

1.链接地址: http://bailian.openjudge.cn/practice/1207/ http://poj.org/problem?id=1207 2.题目: 总时间限制: 1000ms 内存限制: 65536kB 描述 Problems in Computer Science are often classified as belonging to a certain class of problems (e.g., NP, Unsolvable, Recursive). In…

http://poj.org/problem?id=3320 题意:给出一串数字,要求包含所有数字的最短长度. 思路: 哈希一直不是很会用,这道题也是参考了别人的代码,想了很久. #include<iostream> #include<algorithm> #include<string> #include<cstring> using namespace std; ; int n; int len; //开散列法,也就是用链表来存储,所以下面的len是从P…

id=1681">http://poj.org/problem? id=1681 求最少经过的步数使得输入的矩阵全变为y. 思路:高斯消元求出自由变元.然后枚举自由变元,求出最优值. 注意依据自由变元求其它解及求最优值的方法. #include <stdio.h> #include <algorithm> #include <set> #include <map> #include <vector> #include <ma…

水两发去建模,晚饭吃跟没吃似的,吃完没感觉啊. ---------------------------分割线"水过....."------------------------------- POJ 3100 Root of the Problem http://poj.org/problem?id=3100 大意: 给定B和N,求一个数A使得A^N最接近B 太水了..... #include<cstdio> #include<cmath> int main()…

地址 http://poj.org/problem?id=3320 解答 使用双指针 在指针范围内是否达到要求 若不足要求则从右进行拓展  若满足要求则从左缩减区域 代码如下  正确性调整了几次 然后被输入卡TLE卡了很久都没意识到......... #include <iostream> #include <map> #include <set> #include <algorithm> #include <assert.h> #include…

题目传送门 /* 尺取法:先求出不同知识点的总个数tot,然后以获得知识点的个数作为界限, 更新最小值 */ #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <set> #include <map> using namespace std; ; const int INF = 0x3f3f3f3f; int a[MA…