POJ1840 解方程（数的哈希）

给定a1, a2, a3, a4, a5，求符合条件的解（x1, x2, x3, x4, x5）的个数，使得a1*x1^3 + a2*x2^3 + a3*x3^3 + a4*x4^3 + a5*x5^3 = 0。其中-50 <= x1, x2, x3, x4, x5 <= 50且x1-x5都不为0。

x1-x5各有100个可能值，相互的解法是使用5重循环逐一判断，但这样用的时间太长了。因此可以把它们分为两部分（x1, x2）和（x3, x4, x5），求出所有的sum = a1*x1^3 + a2*x2^3，将非负的sum值和负的sum值映射到非负数和负数两个哈希表上，把sum相同的压缩到一个结点，只记录相同的sum的个数。然后求出所有的sum = a3*x3^3 + a4*x4^3 + a5*x5^3，然后分别到非负哈希表和负数哈希表上查找对应的（x1, x2）组合的个数。把所有的结果累加起来就得到最终的答案了。

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

const int N = 10005;
const int H = 9997;

struct Node
{
	long sum;
	int cnt;
	int next;
};
Node nodeP[N], nodeN[N];
int curP, curN;
int hashTableP[H], hashTableN[H];
int a1, a2, a3, a4, a5;
int cube[101];
long long ans;

void initHash()
{
	curP = curN = 0;
	ans = 0;
	for (int i = 0; i < H; ++i) hashTableP[i] = hashTableN[i] = -1;
	for (int i = 0; i <= 100; ++i)
	{
		int t = i - 50;
		cube[i] = t * t * t;
	}
}

void insertHash(int i, int j)
{
	long sum = cube[i] * a1 + cube[j] * a2;
	int h = sum % H;
	int next;
	if (h >= 0)
	{
		next = hashTableP[h];
		while (next != -1)
		{
			if (nodeP[next].sum == sum) 
			{
				++nodeP[next].cnt;
				return;
			}
			next = nodeP[next].next;
		}
		nodeP[curP].cnt = 1;
		nodeP[curP].sum = sum;
		nodeP[curP].next = hashTableP[h];
		hashTableP[h] = curP;
		++curP;
	}
	else
	{
		h = -h;
		next = hashTableN[h];
		while (next != -1)
		{
			if (nodeN[next].sum == sum)
			{
				++nodeN[next].cnt;
				return;
			}
			next = nodeN[next].next;
		}
		nodeN[curN].cnt = 1;
		nodeN[curN].sum = sum;
		nodeN[curN].next = hashTableN[h];
		hashTableN[h] = curN;
		++curN;
	}
}

void getAns(int i, int j, int k)
{
	long sum = cube[i] * a3 + cube[j] * a4 + cube[k] * a5;
	int h = sum % H;
	int next;
	if (h > 0)
	{
		next = hashTableN[h];
		while (next != -1)
		{
			if (nodeN[next].sum + sum == 0)
			{
				ans += nodeN[next].cnt;
				return;
			}
			next = nodeN[next].next;
		}
	}
	else
	{
		h = -h;
		next = hashTableP[h];
		while (next != -1)
		{
			if (nodeP[next].sum + sum == 0)
			{
				ans += nodeP[next].cnt;
				return;
			}
			next = nodeP[next].next;
		}
	}
}

int main()
{
	initHash();
	scanf("%d%d%d%d%d", &a1, &a2, &a3, &a4, &a5);
	for (int i = 0; i <= 100; ++i)
	{
		if (i == 50) continue;
		for (int j = 0; j <= 100; ++j)
		{
			if (j == 50) continue;
			insertHash(i, j);
		}
	}
	for (int i = 0; i <= 100; ++i)
	{
		if (i == 50) continue;
		for (int j = 0; j <= 100; ++j)
		{
			if (j == 50) continue;
			for (int k = 0; k <= 100; ++k)
			{
				if (k == 50) continue;
				getAns(i, j, k);
			}
		}
	}
	printf("%lld\n", ans);
	return 0;
}