HDU 5616 Jam's balance(Jam的天平) Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)   Problem Description - 题目描述 Jim has a balance and N weights. (1≤N≤20) The balance can only tell whether things on different side are t…

题目网址:http://acm.hdu.edu.cn/showproblem.php?pid=5616 题目: Jam's balance Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1810    Accepted Submission(s): 754 Problem Description   Jim has a balance…

题目网址:http://acm.hdu.edu.cn/showproblem.php?pid=5616 题目: Jam's balance Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1810    Accepted Submission(s): 754 Problem Description   Jim has a balance…

Jam's balance Problem Description Jim has a balance and N weights. (1≤N≤20)The balance can only tell whether things on different side are the same weight.Weights can be put on left side or right side arbitrarily.Please tell whether the balance can me…

http://acm.hdu.edu.cn/showproblem.php?pid=5616 思路 题目中蕴含着两种需要计算的重量 1. 从所有的砝码中挑出任意种2.(转换的思想)在天平的两端都挑出这些砝码中的一些,它也可以计算出差值,这个差值也是一种重量    这其实一种转换,看下面这个公式(左端砝码的重量)==(右端砝码的重量+能计算出的重量G)    当第一次选择一个砝码,我们计算出了以第二种方式的一种重量G,那么它也要被记为true,它还可以    继续更新出新的重量,G(新的左端砝码重…

来自官方题解: AC代码: #pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cstring> #include<cmath> #include<math.h> #include<algorithm> #include<queue> #include&l…

背包.dp[i]=1表示i这种差值能被组合出来,差值有负数,所以用sum表示0,0表示-sum,2*sum表示sum. 询问X的时候,只需看dp[sum+X]或者dp[sum-X]是否有一个为1,注意RE. #include<cstdio> #include<cstring> #include<cmath> #include<vector> #include<algorithm> #include<queue> using names…

hdu 5618 Jam's problem again #include <bits/stdc++.h> #define MAXN 100010 using namespace std; int n,k,T,xx; int ans[MAXN],c[MAXN],f[MAXN]; struct Node{ int x,y,z,id; }a[100010],b[100010]; inline int read(){ char ch; bool f=false; int res=0; while (…

Jam's balance Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 370    Accepted Submission(s): 173 Problem Description Jim has a balance and N weights. (1≤N≤20) The balance can only tell whether t…

http://acm.hdu.edu.cn/showproblem.php?pid=5616 题意:有n个物品,每个重量为w[i],有一个天平,你可以把物品放在天平的左边或者右边,接下来m个询问,问是否能用这些物品称出x的重量. 思路:如果只能放一边,那就是很简单的背包DP,现在考虑要放在左右两边,那么只需要逆着来做一遍就好了.因为如果放在左边就不能放在右边了,因此要分开算. 初始状态dp[0][0] = 1. if(dp[i-1][j+w[i]] == 1) dp[i][j] = 1.(j +…

Jim has a balance and N weights. (1≤N≤20) The balance can only tell whether things on different side are the same weight. Weights can be put on left side or right side arbitrarily. Please tell whether the balance can measure an object of weight M. In…

Jam's math problem Problem Description Jam has a math problem. He just learned factorization.He is trying to factorize ax^2+bx+c into the form of pqx^2+(qk+mp)x+km=(px+k)(qx+m).He could only solve the problem in which p,q,m,k are positive numbers.Ple…

Jam's store Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 61    Accepted Submission(s): 18 Problem Description Jam didn't study well,then he go to repair the computer in a store,there are M st…

Jam's problem again Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 640    Accepted Submission(s): 210 Problem Description Jam like to solve the problem which on the 3D-axis,given N(1≤N≤100000)…

题目链接: hdu: http://acm.hdu.edu.cn/showproblem.php?pid=5617 bc(中文):http://bestcoder.hdu.edu.cn/contests/contest_chineseproblem.php?cid=666&pid=1003 题解: 设dp[x1][x2][i]表示第i步时,从(1,1)点走到了(x1,y1),(n,n)点走到了(x2,y2)点的合法的总数. #include<iostream> #include<…

d. Jam有道数学题想向你请教一下,他刚刚学会因式分解比如说,x^2+6x+5=(x+1)(x+5) 就好像形如 ax^2+bx+c => pqx^2+(qk+mp)x+km=(px+k)(qx+m) 但是他很蠢,他只会做p,q,m,kp,q,m,k为正整数的题目 请你帮助他,问可不可以分解 题意就是问一个一元二次方程能不能进行十字相乘的分解? s. 官方题解:第一道题比较简单,可以说是简单的模拟题,我们考虑到a,b,c都是10^9​​的,所以我们决定要把时间复杂度降下来, 对于每一个数,因为…

Jim has a balance and N weights. (1≤N≤20)(1≤N≤20) The balance can only tell whether things on different side are the same weight. Weights can be put on left side or right side arbitrarily. Please tell whether the balance can measure an object of weig…

题意:给你1e5个点(x,y,z),对于每一个点询问有多少个点(x1,y1,z1)满足x1<=x&&y1<=y&&z1<=z 分析:(官方题解奉上)很显然让你找(x,y,z)(x,y,z)都大于别的(x,y,z)(x,y,z),当然厉害的人可以用树套树水一下,但正解写的是CDQ分治,以xx为第一关键字排序,以yy为第二关键字来找,以zz为第三关键字建一个树状数组找就好了,当然等于的情况可以实现前做一下. #include <cstdio> #i…

题意:给你一个字符矩阵,从(1,1)到(n,n)有很多种走法,每一种走法形成一个字符串,问有多少种走法形成的字符串是回文的 分析:(粘贴BC题解) 的是回文串,有人会想到后缀数组自动机马拉车什么的,其实只要求方案数很多,所以我们应该想到动态规划,首先是状态的定义,我们可以想着从(1,1)(1,1)和(n,n)(n,n)开始走,然后走到相遇.所以我们定义状态可以定义为f[x_1][y_1][x_2][y_2]f[x​1​​][y​1​​][x​2​​][y​2​​]表示所对应的两个点(x_1,y_…

题意: 三维坐标,对于1个点,找出有多少个点,3个坐标都比该点小! Sample Input 1 4 10 4 7 10 6 6 8 2 5 7 3 10   Sample Output 1 1 0 0     首先是方法一: 很常见的三维偏序做法,先将所有输入点按要求排序,然后向数据结构插入的时候就可以确定先插入的x值一定比后插入的x小,这样就将三维转化为2维. 2维的话就用树状数组套线段树.根据y建树状数组,根据z建线段树,每一个树状数组的点都对应着一个线段树. #include<cstdi…

方法一:由十字相乘相关理论我们能知道,如果要有p,k,q,m,那么首先要有解,所以b*b-4*a*c要>0,然而因为p,k,q,m是正整数,所以代表x1,x2都是有理数,有理数是什么鬼呢?就是解不带根号,我们知道有求根公式,其中2*a,-b都保证是自然数了,如果根号下b*b-4*a*c也保证是有理数我们就就能保证解是自然数,那么如何保证根号下b*b-4*a*c是有理数呢?那么b*b-4*a*c就是平方数 #pragma comment(linker, "/STACK:1024000000,…

题意:给n个点,求每一个点的满足 x y z 都小于等于它的其他点的个数. 析:三维的,第一维直接排序就好按下标来,第二维按值来,第三维用数状数组维即可. 代码如下: cdq 分治: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #…

惨败,不能再嘲笑别人了,否则自己也会像别人那样倒霉 HDU 5615:http://acm.hdu.edu.cn/showproblem.php?pid=5615 求ax^2+bx+c能否拆成(px+k)(qx+m)的形式 不错的方法,原来的被hack了 #include <cstdio> #include <iostream> #include <cmath> #include <cstring> #include <queue> #inclu…

2017-08-25 20:08:54 writer:pprp 题目简述: • HDU 5616• n个砝码,可以放在天平左右两侧或不放• m次询问,每次询问是否可以测出给定重量• 1 ≤ n ≤ 20• 1 ≤ m ≤ 100 这道题采用枚举的思路的话实现起来还是有点困难的, 要实现的功能是对每个砝码进行处理,加到左边, 加到右边,或者是不加 看了大神的代码,感觉很巧妙, 设置了两个标记数组 vis1[2005], vis2[2005] 一个vis1用来记录当前已经可以实现的重量 另一个vis…

The Balance Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 6652    Accepted Submission(s): 2730 Problem Description Now you are asked to measure a dose of medicine with a balance and a number o…

The Balance Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5706    Accepted Submission(s): 2311 Problem Description Now you are asked to measure a dose of medicine with a balance and a number o…

Problem Description Jam has a math problem. He just learned factorization. He is trying to factorize ax^2+bx+cax​2​​+bx+c into the form of pqx^2+(qk+mp)x+km=(px+k)(qx+m)pqx​2​​+(qk+mp)x+km=(px+k)(qx+m). He could only solve the problem in which p,q,m,…

http://acm.hdu.edu.cn/showproblem.php?pid=5618 题意:…… 思路:和NEUOJ那题一样的.重新写了遍理解了一下,算作处理三维偏序的模板了. #include <cstdio> #include <algorithm> #include <iostream> #include <cstring> using namespace std; #define INF 0x3f3f3f3f #define N 100010…

The Balance Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5956    Accepted Submission(s): 2427 Problem Description Now you are asked to measure a dose of medicine with a balance and a number o…

题意: 现在你被要求用天平和一些砝码来量一剂药.当然,这并不总是可以做到的.所以你应该找出那些不能从范围[1,S]中测量出来的品质.S是所有重量的总质量. 输入一个n,后面有n个数,表示这n个物品的质量 题解: 注意这个题的题干,这个天平只要能制造出来那个质量差,然后这个质量差就不会输出(看懂第二个样例就行) 这道题是母函数的变化版,没学过母函数的可以看一下:母函数 <普通母函数(HDU - 1028 ) && 指数型母函数(hdu1521)> 代码: 1 #include&l…