百练OJ-Eqs

#include<math.h>
#include<iostream>
#include<vector>
#include<cstring>
using namespace std;
vector<int>ranges;
//map<int,short> leftnums;
short leftnums[25000001];
int a1, a2, a3, a4, a5;
//int eqs(int x1, int x2, int x3, int x4, int x5) {
//	return a1*pow(x1,3) + a2 * pow(x2,3) + a3 *pow(x3,3) + a4 * pow(x4,3) + a5 * pow(x5,3);
//}
int solution() {
	int result = 0;
	//for (int a1 = 0; a1 < ranges.size(); a1++) {
	//	for (int a2 = 0; a2 < ranges.size(); a2++) {
	//		for (int a3 = 0; a3 < ranges.size(); a3++) {
	//			for (int a4 = 0; a4 < ranges.size(); a4++) {
	//				for (int a5 = 0; a5 < ranges.size(); a5++) {
	//					if (0 == eqs(ranges[a1], ranges[a2], ranges[a3], ranges[a4], ranges[a5]))
	//						result++;
	//				}
	//			}
	//		}
	//	}
	//}
	for (int x1 = 0; x1 < ranges.size(); x1++) {
		for (int x2 = 0; x2 < ranges.size(); x2++) {
			int num=-(a1*pow(ranges[x1], 3) + a2 * pow(ranges[x2],3));
			if (num < 0) {
				num += 25000000;
			}
			leftnums[num] += 1;
		}
	}
	for (int x3 = 0; x3 < ranges.size(); x3++) {
		for (int x4 = 0; x4 < ranges.size(); x4++) {
			for (int x5 = 0; x5 < ranges.size(); x5++) {
				int num = a3 *pow(ranges[x3], 3) + a4 * pow(ranges[x4], 3) + a5 * pow(ranges[x5], 3);
				if (num < 0) {
					num += 25000000;
				}
				result+=leftnums[num];
			}
		}
	}
	return result;
}
int main()
{
	for (int i = 1; i <= 50; i++) {
		ranges.push_back(i);
		ranges.push_back(-i);
	}
	while (cin >> a1 >> a2 >> a3 >> a4 >> a5) {
		memset(leftnums, 0, sizeof(leftnums));
		cout << solution() << endl;
	}
    return 0;
}