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
首先将等式做一个简单的变换： -(a1*x1^3 + a2*x2^3) = a3*x3^3+a4*x4^3+a5*x5^3 ，然后运用哈希表！
#include <iostream> #include <cstdlib> using namespace std; #define sum 25000001 short a[sum]; int main() { int a1, a2, a3, a4, a5; cin >> a1>> a2>> a3>> a4>> a5; memset(a, 0, sum); for (int x1 = -50; x1<=50; x1++) { if ( !x1 ) continue; for (int x2 = -50; x2<=50; x2++) { if ( !x2 ) continue; int sum1 = (a1*x1*x1*x1 + a2*x2*x2*x2) *(-1); if (sum1 < 0) sum1 = sum1 + 25000000; a[sum1] ++ ; } } int num = 0; for (int x3 = -50; x3<=50; x3++) { if ( !x3 ) continue; for (int x4 = -50; x4<=50; x4++) { if ( !x4 ) continue; for (int x5 = -50; x5<=50; x5++) { if ( !x5 ) continue; int sum2 = a3*x3*x3*x3 + a4*x4*x4*x4 + a5*x5*x5*x5; if (sum2 < 0) sum2 = sum2 + 25000000; if (a[sum2]) num = num + a[sum2]; } } } cout << num << endl; return 0; }