POJ1840(哈希)
大意：
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.
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
求方程解的个数。
分析；将原式变为
a1x1
3+ a2x2
3+ a3x3
3=- a4x4
3- a5x5
3
对左式枚举哈希处理，再计算右式查询。
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#define maxn 1030307
int hashn[maxn], nextn[maxn];
int num[maxn];
using namespace std;
int main()
{
int a, b, c, d, e;
while (~scanf("%d%d%d%d%d", &a, &b, &c, &d, &e))
{
memset(hashn, -1, sizeof(hashn));
int mm = 0;
for (int i = -50; i <= 50; i++)
if (i != 0)
for (int j = -50; j <= 50; j++)
if (j != 0)
for (int k = -50; k <= 50; k++)
if (k != 0)
{
int t = a*i*i*i + b*j*j*j + c*k*k*k;
num[mm] = t;
int key = t%maxn;
key = (key + maxn) % maxn;
nextn[mm] = hashn[key];
hashn[key] = mm;
mm++;
}
int sum = 0;
for (int i = -50; i <= 50; i++)
if (i != 0)
for (int j = -50; j <= 50; j++)
if (j != 0)
{
int t = -e*i*i*i - d*j*j*j;
int key = t%maxn;
key = (key + maxn) % maxn;
int m = hashn[key];
while (m != -1)
{
if (t == num[m])
sum++;
m = nextn[m];
}
}
printf("%d\n", sum);
}
return 0;
}