zoukankan      html  css  js  c++  java
  • poj 1840 -- Eqs

    Eqs
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 11865   Accepted: 5811

    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

    思路:hash水过。

     1 /*======================================================================
     2  *           Author :   kevin
     3  *         Filename :   Eqs.cpp
     4  *       Creat time :   2014-07-24 17:49
     5  *      Description :
     6 ========================================================================*/
     7 #include <iostream>
     8 #include <algorithm>
     9 #include <cstdio>
    10 #include <cstring>
    11 #include <queue>
    12 #include <cmath>
    13 #define clr(a,b) memset(a,b,sizeof(a))
    14 #define M 12500000
    15 using namespace std;
    16 short int ha[2*M+5];
    17 int main(int argc,char *argv[])
    18 {
    19     int a,b,c,d,e;
    20     while(scanf("%d%d%d%d%d",&a,&b,&c,&d,&e)!=EOF){
    21         clr(ha,0);
    22         for(int i = -50; i <= 50; i++){
    23             if(!i) continue;
    24             for(int j = -50; j <= 50; j++){
    25                 if(!j) continue;
    26                 int s = a*i*i*i+b*j*j*j;
    27                 ha[M-s]++;
    28             }
    29         }
    30         int cnt = 0;
    31         for(int i = -50; i <= 50; i++){
    32             if(!i) continue;
    33             for(int j = -50; j <= 50; j++){
    34                 if(!j) continue;
    35                 for(int k = -50; k <= 50; k++){
    36                     if(!k) continue;
    37                     int s = c*i*i*i+d*j*j*j+e*k*k*k;
    38                     s = M+s;
    39                     if(s < 2*M && s > 0)
    40                         cnt += ha[s];
    41                 }
    42             }
    43         }
    44         printf("%d
    ",cnt);
    45     }
    46     return 0;
    47 }
    View Code
  • 相关阅读:
    Java性能权威指南读书笔记--之二
    Java性能权威指南读书笔记--之一
    深入理解JVM-java字节码文件结构剖析(练习解读字节码)
    深入理解JVM-java字节码文件结构剖析(1)
    jvm(5)---垃圾回收(回收算法和垃圾收集器)
    jvm(4)---垃圾回收(哪些对象可以被回收)
    jvm(3)---常用监控工具指令
    jvm(2)---类加载机制
    jvm(1)---java内存结构
    Eureka客户端源码流程梳理
  • 原文地址:https://www.cnblogs.com/ubuntu-kevin/p/3866389.html
Copyright © 2011-2022 走看看