Median Weight Bead Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 3162 Accepted: 1630 Description There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, -, N.…

链接:poj 1975 题意:n个珠子,给定它们之间的重量关系.按重量排序.求确定肯定不排在中间的珠子的个数 分析:由于n为奇数.中间为(n+1)/2,对于某个珠子.若有至少有(n+1)/2个珠子比它重或轻,则它肯定不排在中间 能够将能不能确定的权值初始化为0,能确定重量关系的权值设为1 #include<stdio.h> #include<string.h> int a[110][110]; int main() { int T,n,m,i,j,k,d,x,sum; scanf(…

Median Weight Bead Time Limit: 1000ms Memory Limit: 30000KB This problem will be judged on PKU. Original ID: 197564-bit integer IO format: %lld      Java class name: Main   There are N beads which of the same shape and size, but with different weight…

Description There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, ..., N. Your task is to find the bead whose weight is median (the ((N+1)/2)th among all beads). The foll…

POJ 1979 Red and Black (红与黑) Time Limit: 1000MS    Memory Limit: 30000K Description 题目描述 There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to…

POJ 3268 Silver Cow Party (最短路径) Description One cow from each of N farms (1 ≤ N ≤ 1000) conveniently numbered 1..N is going to attend the big cow party to be held at farm #X (1 ≤ X ≤ N). A total of M (1 ≤ M ≤ 100,000) unidirectional (one-way roads c…

POJ.3087 Shuffle'm Up (模拟) 题意分析 给定两个长度为len的字符串s1和s2, 接着给出一个长度为len*2的字符串s12. 将字符串s1和s2通过一定的变换变成s12,找到变换次数 变换规则如下: 假设s1=12345,s2=67890 变换后的序列 s=6172839405 如果s和s12完全相等那么输出变换次数 如果不完全相等,s的前半部分作为s1,后半部分作为s2,重复上述过程. 直接模拟,注意给出的顺序是从底到上的. 代码总览 #include <cstdio…

POJ.1426 Find The Multiple (BFS) 题意分析 给出一个数字n,求出一个由01组成的十进制数,并且是n的倍数. 思路就是从1开始,枚举下一位,因为下一位只能是0或1,故这个数字只能是1 * 10或者1 * 10 + 1.就按照这种方式枚举,依次放入队列,如果是其的倍数,就输出. 一开始没理解题意,以为是找一个能整除的二进制数,错了半天. 代码总览 #include <cstdio> #include <cstring> #include <algo…

前言:昨天contest4的惨败经历让我懂得要想在ACM领域拿到好成绩,必须要真正的下苦功夫,不能再浪了!暑假还有一半,还有时间!今天找了POJ的分类题库,做了简单题目类型中的图论专题,还剩下二分图和最大流两个子专题没有完成,将在简析(二)中放出. //最短路径 POJ1860:题目链接大意:给定初始货币,给一些货币兑换点,每个货币兑换点可将货币A以E1的汇率和V1的手续费兑换成货币B,和将货币B以E2的汇率和V2的手续费兑换成货币A,给定初始所在货币点,问初始货币能否在一系列兑换后回到初始所在…

Description There are N beads which of the same shape and size, but with different weights. N is an odd number and the beads are labeled as 1, 2, ..., N. Your task is to find the bead whose weight is median (the ((N+1)/2)th among all beads). The foll…