zoukankan      html  css  js  c++  java
  • poj1840Eqs【散列表】

    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∈[-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

    分析:
    把前两个hash一下进行存值
    然后再后面查询有没有出现过即可

    代码:
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <cmath>
     5 using namespace std;
     6 
     7 const int mod = 100007;
     8 
     9 struct Node {
    10     int d;
    11     Node* next;
    12 };
    13 
    14 Node* head[mod + 10];
    15 Node nd[mod + 10];
    16 
    17 int main() {
    18     int a, b, c, d, e;
    19     while(EOF != scanf("%d %d %d %d %d",&a, &b, &c, &d, &e) ) {
    20         memset(head, 0, sizeof(head));
    21         int n_cnt = 0;
    22         for(int i = -50; i <= 50; i++) {
    23             for(int j = -50; j <= 50; j++) {
    24                 if(i == 0 || j == 0) continue;
    25                 int num = a * i * i * i + b *j * j * j;
    26                 int xx = num > 0 ? num : -num;
    27                 int p = xx % mod;
    28                 Node*pt = head[p];
    29                 while(pt) {
    30                     pt = pt -> next;
    31                 }
    32                 nd[n_cnt].d = num;
    33                 nd[n_cnt].next = head[p];
    34                 head[p] = &nd[n_cnt++];
    35             }
    36         }
    37         int ans = 0;
    38         for(int i = -50; i <= 50; i++) {
    39             for(int j = -50; j <= 50; j++) {
    40                 for(int k = -50; k <= 50; k++) {
    41                     if(i == 0 || j == 0 || k == 0) continue;
    42                     int num = c * i * i * i + d * j * j * j + e * k * k * k;
    43                     num = - num;
    44                     int xx = num > 0 ? num : - num;
    45                     int p = xx % mod;
    46                     Node * pt = head[p];
    47                     while(pt) {
    48                         if(pt -> d == num) ans++;
    49                         pt = pt -> next;
    50                     }
    51                 }
    52             }
    53         }
    54         printf("%d
    ", ans);
    55     }
    56     return 0;
    57 }
    动态内存申请
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 using namespace std;
     5 
     6 const int mod = 10007;
     7 
     8 struct Node {
     9     int to;
    10     int next;
    11 }e[mod + 10];
    12 
    13 int head[mod + 10];
    14 
    15 int tot;
    16 void add(int u, int v) {
    17     e[tot].to = v;
    18     e[tot].next = head[u];
    19     head[u] = tot++;
    20 }
    21 
    22 int Find(int p, int num) {
    23     int cnt = 0;
    24     for(int i = head[p]; i; i = e[i].next) {
    25         if(e[i].to == num) cnt++;
    26     }
    27     return cnt;
    28 }
    29 
    30 int Fabs(int x) {
    31     return x > 0 ? x : - x;
    32 }
    33 
    34 int main() {
    35     int a, b, c, d, e;
    36     while(EOF != scanf("%d %d %d %d %d",&a, &b, &c, &d, &e) ) {
    37         memset(head, 0, sizeof(head));
    38         tot = 1;
    39         for(int i = -50; i <= 50; i++) {
    40             for(int j = -50; j <= 50; j++) {
    41                 if(i == 0 || j == 0) continue;
    42                 int num = a * i * i * i + b * j * j * j;
    43                 int p = Fabs(num) % mod;
    44                 add(p, num);
    45             }
    46         }
    47         int ans = 0;
    48         for(int i = -50; i <= 50; i++) {
    49             for(int j = -50; j <= 50; j++) {
    50                 for(int k = -50; k <= 50; k++) {
    51                     if(i == 0 || j == 0 || k == 0) continue;
    52                     int num = c * i * i * i + d * j * j * j + e * k * k * k;
    53                     num = - num;
    54                     int p = Fabs(num) % mod;
    55                     ans += Find(p, num);
    56                 }
    57             }
    58         }
    59         printf("%d
    ",ans);
    60     }
    61     return 0;
    62 }
    数组模拟


  • 相关阅读:
    call_user_func和call_user_func_array的区别
    25行实现文件上传功能(PHP)
    PHP 结合MYSQL简单的实现了Todo List 功能
    向ASP.NET自定义控件中嵌入CSS资源
    模态子窗口不执行page_load
    oracle一次插入多条数据
    转载:数据库表结构设计方法及原则
    Js中的window.parent ,window.top,window.self
    在一个JS文件中包含中文字符串通过innerHTML输出后中文乱码
    转载:数据库表结构设计方法及原则
  • 原文地址:https://www.cnblogs.com/zhanzhao/p/3991368.html
Copyright © 2011-2022 走看看