POJ 1840 - Eqs, HRBUST 1013 - Eqs [hash]

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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.

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Output

The output will contain on the first line the number of the solutions for the given equation.

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Sample Input

37 29 41 43 47

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Sample Output

654

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The meaning of problems

Entry $5$ number $(A1, a2, a3, a4, a5) ∈ [-50,50]$ , seeking to satisfy $a1 *$ $x1$ ³ $+ A2 *$ $x2$ ³ $+ A3 *$ $x3$ ³ $+ A4 *$ $x4$ ³ $+ A5 *$ $x5$ ³ $=0$ The number of the polynomial $(x1,x2,x3,x4,x5)∈[-50,50]$。

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answer

• $5$ layer solution is certainly the time complexity of the cycle of violence is not allowed, so the converted $a1 *$x1³ $+ A2 *$x2³ $= - (a3 *$x3³ $+ A4 *$x4³ $+ A5 *$x5³ $)$
• As can be seen, the upper limit of the range to satisfy the equation $50*50$ ^{. 3}, since the index can not be negative, increasing the unified array subscript $mid$
• Then the water is too violent ~~~

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AC-Code

#inlcude<bits/stdc++.h>
using namespace std;

const int maxn = 12500000 << 1;
short h[maxn];
int main() {
	const int mid = 12500000;
	int a, b, c, d, e;
	while (cin >> a >> b >> c >> d >> e) {
		memset(h, 0, sizeof h);
		int ans = 0;
		for (int i = -50; i <= 50; ++i) {
			if (!i)	continue;
			for (int j = -50; j <= 50; ++j) {
				if (!j)	continue;
				++h[a * i * i * i + b * j * j * j + mid];
			}
		}
		for (int i = -50; i <= 50; ++i) {
			if (!i)	continue;
			for (int j = -50; j <= 50; ++j) {
				if (!j)	continue;
				for (int k = -50; k <= 50; ++k) {
					if (!k)	continue;
					if (-(c * i * i * i + d * j * j * j + e * k * k * k) + mid < maxn
						and -(c * i * i * i + d * j * j * j + e * k * k * k) + mid >= 0)
						ans += h[-(c * i * i * i + d * j * j * j + e * k * k * k) + mid];
				}
			}
		}
		cout << ans << endl;
	}

	return 0;
}