zoukankan      html  css  js  c++  java
  • Eqs 哈希

                                                       Eqs

    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∈[-50,50], xi != 0, any i∈{1,2,3,4,5}. 

    Determine how many solutions satisfy the given equation. 

    Input

    The only line of input contains the 5 coefficients a1, a2, a3, a4, a5, separated by blanks.

    Output

    The output will contain on the first line the number of the solutions for the given equation.

    Sample Input

    37 29 41 43 47

    Sample Output

    654

    Source

    #include<stdio.h>
    #include<string.h>
    short a[12500001], b[12500001];
    int main()
    {
        int a1, a2, a3, a4, a5, ans=0, i, j;
        scanf("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5);
        memset(a, 0, sizeof(a));
        memset(b, 0, sizeof(b));
        for(i=-50; i<=50; i++)
            for(j=-50; j<=50; j++)
            {
                if(i==0 || j==0)
                    continue;
                int sum = a1*i*i*i + a2*j*j*j;
                if(sum <= 0)
                    a[-sum]++;
                else
                    b[sum]++;
            }
        for(i=-50; i<=50; i++)
            for(j=-50; j<=50; j++)
                for(int k=-50; k<=50; k++)
                {
                    if(i==0 || j==0 || k==0)
                        continue;
                    int sum = a3*i*i*i + a4*j*j*j + a5*k*k*k;
                    if(-12500000<=sum && sum<=12500000)
                    {
                        if(sum < 0)
                        {
                            if(b[-sum] != 0)
                                ans += b[-sum];
                        }
                        else
                        {
                            if(a[sum] != 0)
                                ans += a[sum];
                        }
                    }
                }
        printf("%d
    ", ans);
        return 0;
    }


  • 相关阅读:
    Dubbo源码分析系列---服务的发布
    Dubbo源码分析系列---扩展点加载
    Jdk动态代理和CGLIB动态代理大比拼
    定时任务的一些思路
    技术人的职业发展
    2017面试碎碎念
    Tiny Mapper
    RabbitMQ 简介
    Load Test Analyzer Overview
    2015 如期而至,你好
  • 原文地址:https://www.cnblogs.com/Genesis2018/p/9079862.html
Copyright © 2011-2022 走看看