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poj 1840 Eqs

2016-08-03 15:53 465 查看

Eqs

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 15314		Accepted: 7518

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

Source

Romania OI 2002

提示

题意：

我们给出一个方程：

a1x1^3+a2x2^3+a3x3^3+a4x4^3+a5x5^3=0。（-50<=ai<=50，-50<=xi<0∪0<xi<=50）

求有几种情况满足方程？

思路：

a1x1^3+a2x2^3=-（a3x3^3+a4x4^3+a5x5^3）。

先以a1x1^3+a2x2^3的所有值全部打表，注意是全部，50^4+50^4=12500000,连上负数是25000000，并且我们记录的是这个值产生的次数而不是存不存在。

之后用-（a3x3^3+a4x4^3+a5x5^3）去做比较，相同就加。

示例程序

Source Code

Problem: 1840		Code Length: 1147B
Memory: 49356K		Time: 516MS
Language: GCC		Result: Accepted
#include <stdio.h>
#include <string.h>
short a[25000001];
int main()
{
int a1,a2,a3,a4,a5,x1,x2,x3,x4,x5,y,num=0;
memset(a,0,sizeof(a));
scanf("%d %d %d %d %d",&a1,&a2,&a3,&a4,&a5);
for(x1=-50;x1<=50;x1++)
{
if(x1==0)
{
x1++;
}
for(x2=-50;x2<=50;x2++)
{
if(x2==0)
{
x2++;
}
y=a1*x1*x1*x1+a2*x2*x2*x2;
if(y<0)
{
y=y+25000000;			//大于12500000的就是负数了，这里是为了数组下标好做统计
}
a[y]++;
}
}
for(x3=-50;x3<=50;x3++)
{
if(x3==0)
{
x3++;
}
for(x4=-50;x4<=50;x4++)
{
if(x4==0)
{
x4++;
}
for(x5=-50;x5<=50;x5++)
{
if(x5==0)
{
x5++;
}
y=(-1)*(a3*x3*x3*x3+a4*x4*x4*x4+a5*x5*x5*x5);
if(y<0)				//同理
{
y=y+25000000;
}
num=num+a[y];
}
}
}
printf("%d",num);
return 0;
}

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