Eqs
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 11693 | Accepted: 5722 |
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.
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
654
题意:问有多少组x1-x5使得a1x13+ a2x23+ a3x33+ a4x43+ a5x53=0。
思路:分成前后两部分,注意数据范围,50的4次方*3<2*10的7次方。
AC代码如下:
#include<cstdio>
#include<cstring>
using namespace std;
short hash[40000000];
int main()
{ int a1,a2,a3,a4,a5,x1,x2,x3,x4,x5,ans=0;
scanf("%d%d%d%d%d",&a1,&a2,&a3,&a4,&a5);
for(x1=-50;x1<=50;x1++)
if(x1!=0)
for(x2=-50;x2<=50;x2++)
if(x2!=0)
hash[(a1*x1*x1*x1+a2*x2*x2*x2)*(-1)+20000000]++;
for(x3=-50;x3<=50;x3++)
if(x3!=0)
for(x4=-50;x4<=50;x4++)
if(x4!=0)
for(x5=-50;x5<=50;x5++)
if(x5!=0)
ans+=hash[a3*x3*x3*x3+a4*x4*x4*x4+a5*x5*x5*x5+20000000];
printf("%d\n",ans);
}