TOPCODE/SRM567 DIVII 250、500PT

Problem Statement
	The Ninja Turtles often battle the Foot Clan ninjas. The Turtles celebrate each victory with a pizza party. The amount of pizza they eat depends on the number of opponents they have defeated. Denote the number of defeated opponents as N. Three of the four Turtles have a moderate appetite and only consume floor(N / K) pizzas each. The fourth Turtle is always hungry and eats floor(N / 3) pizzas. You are given ints P and K, where P is the total number of pizzas the Turtles ate after a battle. If there exists at least one value of N such that after defeating N opponents the Turtles would eat exactlyP pizzas at the party, return the smallest such N. Otherwise, return -1.
Definition
	Class: NinjaTurtles Method: countOpponents Parameters: int, int Returns: int Method signature: int countOpponents(int P, int K) (be sure your method is public)
Notes
-	floor(X) is equal to the largest integer which is less or equal to X.
Constraints
-	P will be between 1 and 1,000,000, inclusive.
-	K will be between 4 and 100, inclusive.
Examples
0)
	5 4 Returns: 6 If the Turtles defeated 6 opponents, three of the four Turtles would eat floor(6 / 4) = 1 pizza each and the fourth one would eat floor(6 / 3) = 2 pizzas, which makes 5 pizzas in total. Note that you always have to return the smallest possible N. For example, in this scenario for N = 7 the Turtles would also eat 5 pizzas, but 7 is not a correct return value, because 6 is less than 7.
1)
	1 4 Returns: 3 After a fight with three opponents, only the hungry Turtle would eat a pizza.
2)
	13 6 Returns: -1 There is no value of N such that if the Turtles battle N opponents, they eat exactly 13 pizzas forK = 6.
3)
	13 17 Returns: 30 For K = 17, after defeating 30 opponents the Turtles will eat 13 pizzas in total.
4)
	122 21 Returns: 258

class NinjaTurtles
{
 public:int countOpponents(int P, int K)
 {
  int total=0;
  for(int i=1;i<5000000;i++)
  {
   total = 3*(i/K) + i/3;
   if ( total == P)
   {
    return i;
   }
  }
  return -1;
 }
};

此题运用穷举法。。不过注意去i的范围。根据P、K最大值，去300 0000即可。

Problem Statement
	Consider the function SSR (Squared Sum of square Roots) defined on two positive integer parameters: SSR(A, B) = (sqrt(A)+sqrt(B))^2. We are interested in the cases when the value of the function is also an integer. Given ints N and M, return the number of ordered pairs (A, B) such that 1 <= A <=N, 1 <= B <=M and SSR(A, B) is an integer.
Definition
	Class: TheSquareRootDilemma Method: countPairs Parameters: int, int Returns: int Method signature: int countPairs(int N, int M) (be sure your method is public)
Notes
-	The answer to the problem is guaranteed to fit into signed 32-bit integer type under the given constraints.
Constraints
-	N will be between 1 and 77,777, inclusive.
-	M will be between 1 and 77,777, inclusive.
Examples
0)
	2 2 Returns: 2 Out of the four possible pairs (A, B), only two yield an integer result: SSR(1, 1) = 4 and SSR(2, 2) = 8. On the other hand, SSR(1, 2) = SSR(2, 1) = 3+2*sqrt(2), which is not an integer.
1)
	10 1 Returns: 3 SSR(1, 1), SSR(4, 1) and SSR(9, 1) are integers.
2)
	3 8 Returns: 5 The valid pairs are (1, 1), (1, 4), (2, 2), (2, 8) and (3, 3).
3)
	100 100 Returns: 310

class TheSquareRootDilemma
{
bool DivFnd(const vector<int>& arr,const int& n,int start,int end)
{
int tp = (start + end)/2;
if ( (start+1 == end) && (arr[tp] != n))
return false;
if ( arr[tp] == n)
return true;
else
if( arr[tp] < n)
return DivFnd(arr,n,tp,end);
else
return DivFnd(arr,n,start,tp);
return false;
}
public:int countPairs(int N, int M)
{
int max = MAX(N,M);
int min = MIN(N,M);
int cnt = min;
int tpx = (int)sqrt(max);
int i,j,k;
int tpi;
int temp;
vector<int> BaseNum(tpx);
for(i=0;i<tpx;i++)
BaseNum[i] = (i+1)*(i+1);
for(i=1;i<=min;i++)
{
if(DivFnd(BaseNum,i,0,tpx))
{
cnt += (tpx-1);
continue;
}
tpi = i;
for(k = tpx-1;k>0;k--)
{
if (( tpi > BaseNum[k]) && ( tpi%BaseNum[k] ==0))
tpi = tpi/BaseNum[k];
}
for(j = 1;j<tpx;j++)
{
temp = tpi*BaseNum[j];
if( temp > i)
{
if( temp <= max)
{
if ( temp > min)
cnt++;
else
cnt += 2;
}
else
break;
}
}
}
return cnt;
}
};

小弟新手，大神老鸟路过无视即可。

 说下这道题：

这道题题目问题很简单，从N，M中找到数对(a,b),使得a+b+2*sqrt(a*b)为一个整数。其实就是要sqrt(a*b)为一个整数。

由于sqrt()返回一个浮点数，并且直接复制给一个整数i,此时会丢掉小数部分（存在这个问题100 0000和99 9999(这里会进位)开方都会得到1000，而不是1000和999）。再判断一下i*i是不是等于a*b，等于，则说明(a,b)构成数对。

那个程序就是这样：

for(int i=i;i<=N;i++)

for(int j=1;j<=M;j++)

{

temp = sqrt(a*b);

if ( temp*temp == a*b)

cnt++;

}

或者将算法再优化下：

int max = MAX(N,M);

int min = MIN(N,M);

cnt = min;

for(int i=1;i<=N;i++)

  for(int j=i+1;i<=M;j++)

{

temp = sqrt(a*b);

if ( temp*temp == a*b)

{

if( j > N)

cnt++;

else

cnt += 2;

}

}

或者是生成一个int数组，保存从1到MAX(N,M)平方数，然后将(a*b)值查找。

使用上述几种方法，还要考虑int数据不够存放问题，输入77777,77777。理论上都可以。。不过时效差。就第一种而言，输入最大值，循环次数：77777*77777。能提交得分，但过不了system test.。

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上面贴的代码处理思路：

找到n,m中的最大和最小值，先考虑：

1.有相同数据区域。即[1,min] 那有min对

2.在这个区域内，找到数自身是一个数的平方数。sqrt(min)是最大个数 ，可能是sqrt(min)-1个。将1-sqrt(min) or sqrt(min) -1 数据平方后存入数组。便于后续查找，不用每次去判断(可以采用上面说的判断方法).

3.若a 属于这个数组。如：a*16,由于16为4的平方，那么a必须也为这个数组中的一个数。那直接计算：sqrt(min) -1 (减一是因为第一点已经考虑2个数相同）

4.若a 是这个数组中一个数的倍数，则找到这个数，取得倍数。eg:a*18,18为9的2倍，那只需考虑和数字2情况。

不符合3、4的数，是不能够被开方的。

另外，上述代码实现时，(这里不好叙述，直接举例子）

i from 1 to min

eg: i = 2,  直接和它的1 4 9..等能被开方的倍数相乘去判断，在不在最大值范围内。在则累计。累计时，如果大于min,则累计1.否则累计2.(eg: N = 10 ,M = 20,(2,18)可以组成一对，但不存在(18,2).以为不在范围内）

eg: i = 18,先将18除以能被开方的最大倍数，这里处理后是2（这里2已经在前面处理过，可以在前面开一个数组来存放，供后面直接赋值），但要从大于9倍开始，因为9倍以下的数据被累计过了。(在i等于8时累计过了)

所以，只需考虑 i from to min (>min后的数据只有1一个，在前面已经被累计了)

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虽然这次题目比较简单，但个人尝试了多种方法后，感觉收获颇多。

欢迎拍砖！！！