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…

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…

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() {…

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…

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,…

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…

  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…

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…

[题目链接] http://poj.org/problem?id=2785 [题目大意] 给出四个数组,从每个数组中选出一个数,使得四个数相加为0,求方案数 [题解] 将a+b存入哈希表,反查-c-d [代码] #include <cstdio> #include <algorithm> #include <cstring> using namespace std; const int N=4500,mod=1<<22; int n,a[N],b[N],c[…

题意:每片雪花有六瓣,给出n片雪花,六瓣花瓣的长度按顺时针或逆时针给出,判断其中有没有相同的雪花(六瓣花瓣的长度相同) 思路:如果直接遍历会超时,我试过.这里要用哈希表,哈希表的关键码key用六瓣花瓣的长度的和取余一个数得到,表中为雪花的存储位置address(即在snowflakes数组中的位置) 代码: #include<iostream> #include<vector> using namespace std; const int maxn=100000+100;//雪花最…