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 >
using namespace 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);
return 0 ;
}