HDU 4038 无题

/********************************************************************************
	倒不是什么太难的题，主要是细节太阴险了～具体见注释～
	最后学校排排名31～这几场网络赛都相当给力啊～今年挺有戏的～
********************************************************************************/
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <utility>
#include <cstdio>
#include <memory>
#include <string>
#include <vector>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
using namespace std;

#define lowbit(x) ( (x) & ( (x) ^ (x - 1) ) )

//typedef long long LL;
typedef __int64 LL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
typedef map<int, int>::iterator MI;
typedef vector<int>::iterator VI;
typedef list<LL>::iterator LI;
typedef set<int>::iterator SI;

const int INF_INT = 0x3f3f3f3f;
const LL INF_LL = 0x7fffffffffffffff;
const double oo = 10e9;
const double eps = 10e-7;
const double PI = acos(-1.0);

const int MAXN = 100004;
const LL MOD = 1000000007;

int test, n;
LL m, arr[MAXN];

inline LL llabs(LL x)
{
	return x > 0 ? x : -x;
}
LL pow_mod(LL a, LL k)
{
	LL r = 1;
	while (k)
	{
		if (k & 0x1)
		{
			r = (r * a) % MOD;
		}
		a = (a * a) % MOD;
		k >>= 1;
	}
	return r;
}
void ace()
{
	int cas = 1;
	int ncnt, max_id, min_id;
	LL buf;
	LL max_neg, min_pos;
	LL two, tri, four;
	for (scanf("%d", &test); test--; ++cas)
	{
		scanf("%d %I64d", &n, &m);
		ncnt = 0;
		max_neg = -INF_INT;
		for (int i = 0; i < n; ++i)
		{
			scanf("%I64d", arr + i);
			if (arr[i] < 0)
			{
				++ncnt;
				if (max_neg < arr[i])
				{
					max_id = i;
					max_neg = arr[i];
				}
			}
		}
		if (ncnt & 0x1)					//如果有奇数个负数的话，尽量把最大的负数转化为正数
		{
			if (m > llabs( arr[max_id] ))
			{
				m -= llabs( arr[max_id] );
				arr[max_id] = 0;
			}
			else
			{
				arr[max_id] += m;
				m = 0;
			}
		}
		if (m > 0)						//如果有0存在的话，先尽量把0加为1
		{
			for (int i = 0; i < n && m > 0; ++i)
			{
				if (0 == arr[i])
				{
					++arr[i];
					--m;
				}
			}
		}
		if (m > 0)						//如果有1存在的话，尽量把1加为2
		{
			for (int i = 0; i < n && m > 0; ++i)
			{
				if (1 == arr[i])
				{
					++arr[i];
					--m;
				}
			}
		}
		if (m > 0)						//如果有2存在的话，尽量把2加为3
		{
			for (int i = 0; i < n && m > 0; ++i)
			{
				if (2 == arr[i])
				{
					++arr[i];
					--m;
				}
			}
		}
		//开始变了，不能老是加最小的～不信自己列几个数试试～
		two = tri = four = 0;
		if (m > 0)						//还剩下的操作尽量去添3
		{
			min_pos = INF_LL;
			for (int i = 0; i < n; ++i)
			{
				if (arr[i] > 0 && arr[i] < min_pos)		//在给定的集合中找最小的正数，显然应该>=3
				{
					min_id = i;
					min_pos = arr[i];
				}
			}
			tri = m / 3;
			if (1 == m % 3)				//如果还剩下1个操作，那么考虑将整个集合中最小的数+1
			{							//这个地方阴险无比啊
				if (tri > 0)			//如果有3的话，显然把一个3->4
				{
					--tri;
					++four;
				}
				else					//否则把集合中最小的元素+1
				{
					++arr[min_id];
				}
			}
			else if (2 == m % 3)		//如果还剩下2个操作，那么就添一个2上去
			{
				++two;
			}
		}
		buf = 1;
		for (int i = 0; i < n; ++i)
		{
			buf = (buf * arr[i]) % MOD;
		}
		buf = (buf * pow_mod(2, two)) % MOD;
		buf = (buf * pow_mod(3, tri)) % MOD;	//tri必然很大～当然二分幂～
		buf = (buf * pow_mod(4, four))  % MOD;
		printf("Case %d: %I64d\n", cas, buf);
	}
	return ;
}
int main()
{
	ace();
	return 0;
}