POJ 1840 Eqs

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poj 1840 Eqs （hash）

题目:http://poj.org/problem?id=1840 题解:http://blog.csdn.net/lyy289065406/article/details/6647387 小优姐讲的很好了 #include<cstdio> #include<string> #include<iostream> #include<cstring> #include<map> using namespace std; ]; int main() {…

POJ 1840 Eqs 解方程式, 水题难度:0

题目 http://poj.org/problem?id=1840 题意给与数组a[5],其中-50<=a[i]<=50,0<=i<5,求有多少组不同的x[5],使得a[0] * pow(x[0], 3) + a[1] * pow(x[1], 3) + a[2] * pow(x[2], 3) + a[3] * pow(x[3], 3) + a[4] * pow(x[4], 3)==0 其中x[i]满足-50<=x[i]<=50,0<=i<5 思路该等式…

Eqs Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 15010 Accepted: 7366 Description Consider equations having the following form: a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 The coefficients are given integers from the interval [-50,50]. I…

poj 1840 Eqs 【解五元方程+分治+枚举打表+二分查找所有key 】

Eqs Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 13955 Accepted: 6851 Description Consider equations having the following form: a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 The coefficients are given integers from the interval [-50,50]. It i…

POJ 1840 Eqs(hash)

题意输入a1,a2,a3,a4,a5 求有多少种不同的x1,x2,x3,x4,x5序列使得等式成立 a,x取值在-50到50之间直接暴力的话肯定会超时的 100的五次方 10e了都然后能够考虑将等式变一下形把a1*x1^3+a2*x2^3移到右边也就是-(a1*x1^3+a2^x2^3)=a3*x3^3+a4*x4^3+a5*x5^3 考虑到a1*x1^3+a2^x2^3的最大值50*50^3+50*50^3=12500000 这个数并不大能够开这么大…

POJ 1840 Eqs 二分+map/hash

Description Consider equations having the following form: a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 The coefficients are given integers from the interval [-50,50]. It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-50,…

POJ 1840 Eqs 暴力

Description Consider equations having the following form: a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 The coefficients are given integers from the interval [-50,50]. It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈[-5…

POJ 1840 Eqs（乱搞）题解

思路:这题好像以前有类似的讲过,我们把等式移一下,变成 -(a1*x1^3 + a2*x2^3)== a3*x3^3 + a4*x4^3 + a5*x5^3,那么我们只要先预处理求出左边的答案,然后再找右边是否也能得到就行了,暴力的复杂度从O(n^5)降为O(n^3 + n^2).因为左式范围-12500000~12500000,所以至少开12500000 * 2的空间,用int会爆,这里用short.如果小于0要加25000000,这样不会有重复的答案,算是简单的hash? 代码: #incl…

POJ 1840：Eqs 哈希求解五元方程

Eqs Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 14169 Accepted: 6972 Description Consider equations having the following form: a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 The coefficients are given integers from the interval [-50,50]. I…

Eqs - poj 1840(hash)

题意:对于方程:a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 ,有xi∈[-50,50], xi != 0, any i∈{1,2,3,4,5}. 现在给出a1,a2,a3,a4,a5的值,求出满足上面方程的解有多少个. 思路:hash的应用.暴力枚举的话会达到100^5,明显会超时.所以将方程分成-(a1x13+ a2x23 )和 a3x33+a4x43+ a5x53 两部分,若这两部分相等,则为方程的一个解. #include<stdio.h> #include&…

POJ 1840：Eqs

Description Consider equations having the following form: a1x13+ a2x23+ a3x33+ a4x43+ a5x53= The coefficients are given integers ,]. It ,], xi != , any i∈{,,,,}. Determine how many solutions satisfy the given equation. Input The only line of input co…

POJ 1840 HASH

题目链接:http://poj.org/problem?id=1840 题意:公式a1x1^3+ a2x2^3+ a3x3^3+ a4x4^3+ a5x5^3=0,现在给定a1~a5,求有多少个(x1~x5)的组合使得公式成立.并且(x1~x5)取值再[-50,50]且不能为0 思路:因为x的值范围比较小,只有100.所以可以先求出 a1x1^3+a2x2^3+a3x3^3. 然后在求 (-1)*(a4x4^3+a5x5^3)从前面的所得的Hash表进行二分查找. #include<iostre…

POJ 1840 Eps 解题报告（哈希）

a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0,xi∈[-50,50],且xi!=0.让我们求所有解的可能. 首先,如果暴力判断的话,每个x的取值有100种可能,100^5肯定会超时. 我们可以枚举x1,x2的值,并且记录下来.再枚举x3,x4,x5的值.如果发现有互为相反数的,说明有一个解存在.复杂度却大大降低了. 当然,我们可以只处理正数的情况.如果存在一组解,x1,x2,x3,x4,x5,那么容易证明-x1,-x2,-x3,-x4,-x5也是一组解. 我们只记录a1…

POJ 1840 Brainman（逆序对数）

题目链接:http://poj.org/problem?id=1804 题意:给定一个序列a[],每次只允许交换相邻两个数,最少要交换多少次才能把它变成非递降序列. 思路:题目就是要求逆序对数,我们知道,求逆序对最典型的方法就是树状数组,但是还有一种方法就是Merge_sort(),即归并排序.实际上归并排序的交换次数就是这个数组的逆序对个数,归并排序是将数列a[l,h]分成两半a[l,mid]和a[mid+1,h]分别进行归并排序,然后再将这两半合并起来.在合并的过程中(设l<=i<=mid…

poj 1840 枚举

Eqs Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 13967 Accepted: 6858 Description Consider equations having the following form: a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 The coefficients are given integers from the interval [-50,50]. It i…

poj 1840 哈希

Eqs Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 14093 Accepted: 6927 Description Consider equations having the following form: a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 The coefficients are given integers from the interval [-50,50]. It i…

poj 1840 暴力+标记

Description Consider equations having the following form: a1x1 3+ a2x2 3+ a3x3 3+ a4x4 3+ a5x5 3=0 The coefficients are given integers from the interval [-50,50]. It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈…

poj 1840(五元三次方程组）

Description Consider equations having the following form: a1x1 3+ a2x2 3+ a3x3 3+ a4x4 3+ a5x5 3=0 The coefficients are given integers from the interval [-50,50]. It is consider a solution a system (x1, x2, x3, x4, x5) that verifies the equation, xi∈…

poj很好很有层次感（转）

OJ上的一些水题(可用来练手和增加自信) (POJ 3299,POJ 2159,POJ 2739,POJ 1083,POJ 2262,POJ 1503,POJ 3006,POJ 2255,POJ 3094) 初期: 一.基本算法: 枚举. (POJ 1753,POJ 2965) 贪心(POJ 1328,POJ 2109,POJ 2586) 递归和分治法. 递推. 构造法.(POJ 3295) 模拟法.(POJ 1068,POJ 2632,POJ 1573,POJ 2993,POJ 2996) 二…

POJ题目分类推荐（很好很有层次感）

著名题单,最初来源不详.直接来源:http://blog.csdn.net/a1dark/article/details/11714009 OJ上的一些水题(可用来练手和增加自信) (POJ 3299,POJ 2159,POJ 2739,POJ 1083,POJ 2262,POJ 1503,POJ 3006,POJ 2255,POJ 3094) 初期: 一.基本算法: 枚举. (POJ 1753,POJ 2965) 贪心(POJ 1328,POJ 2109,POJ 2586) 递归和分治法. 递…

一些简单二分题,简单的hash,H(i),字符串题

说在前面: 题是乱七八糟的. 几个二分的题. (但是我的做法不一定是二分,有些裸暴力. 1. Equations HDU - 1496 输入a,b,c,d问你这个方程有多少解.a*x1^2+b*x2^2+c*x3^2+d*x4^2=0a,b,c,d属于[-50, 50] x1,x2,x3,x4属于[-100,100]且Xi不等于零解法很简单, 把式子拆两边,拆成a*x1^2 + b*x2^2 = -(c*x3^2 + d*x4^2)的形式,然后暴力枚举就行了. // 注意如果只枚举正数开始枚…

【POJ】1840：Eqs【哈希表】

Eqs Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 18299 Accepted: 8933 Description Consider equations having the following form: a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0 The coefficients are given integers from the interval [-50,50]. It i…

POJ 3349&&3274&&2151&&1840&&2002&&2503

(今天兴致大发学了Markdown,第一篇博客) 这次的主要都是hash的题目(当然这就意味这可以用map) hash的方式也有很多: 普通hash hash挂链双hash以及自然溢出等当然我还是喜欢挂链的(主要是精准) 下面开始看题目 3349 题意很简单,给出一片雪花的信息(六个角) 如果两片雪花相同则它们从某一点开始顺时针或逆时针数字相同这...... 直接hash跑一边即可主意挂链,要不然很容易WA CODE #include<cstdio> #include<cstri…

POJ 题目分类（转载）

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…