poj1840

题意:给出ai(i=1~5),求a1 * x1^3+ a2 * x2^3+ a3 * x3^3+ a4 * x4^3+ a5 * x5^3=0在-50到50之间的x的解的个数

分析:用map,二重循环枚举x1,x2计算结果在map中对应位++

三重循环枚举x3,x4,x5计算结看map中对应的相反数的个数。

View Code
#include <iostream>
#include
<cstdio>
#include
<cstdlib>
#include
<cstring>
#include
<map>
usingnamespace std;

int a[6];
map
<int, int> h;

int main()
{
//freopen("t.txt", "r", stdin);
for (int i =1; i <6; i++)
scanf(
"%d", &a[i]);
for (int i =-50; i <=50; i++)
for (int j =-50; j <=50; j++)
{
if (i ==0|| j ==0)
continue;
int temp = i * i * i * a[1] + j * j * j * a[2];
if (h.find(temp) == h.end())
h[temp]
=1;
else
h[temp]
++;
}
int ans =0;
for (int i =-50; i <=50; i++)
for (int j =-50; j <=50; j++)
for (int k =-50; k <=50; k++)
{
if (i ==0|| j ==0|| k ==0)
continue;
int temp = i * i * i * a[3] + j * j * j * a[4] + k * k * k
* a[5];
if (h.find(0- temp) == h.end())
continue;
ans
+= h[0- temp];
}
printf(
"%d\n", ans);
return0;
}
原文地址:https://www.cnblogs.com/rainydays/p/2076875.html