TOPCODE/SRM567 DIVII 250、500PT

最新推荐文章于 2021-03-19 02:46:09 发布

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最新推荐文章于 2021-03-19 02:46:09 发布

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分类专栏： TOPCODER/SRM567 DIVII 文章标签： 250-500pt 567 SRM TOPCODER

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本文探讨了Ninja Turtles在战胜对手后如何通过数学逻辑计算出最合适的派数量，确保派对的完美进行。文章详细介绍了算法求解过程，包括如何根据已知的总派数量和每只乌龟的食量，找出乌龟们在战胜一定数量的对手后所消耗的派数，从而得出最小可能的对手数量。

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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

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.

Returns: 3

After a fight with three opponents, only the hungry Turtle would eat a pizza.

Returns: -1

There is no value of N such that if the Turtles battle N opponents, they eat exactly 13 pizzas forK = 6.

Returns: 30

For K = 17, after defeating 30 opponents the Turtles will eat 13 pizzas in total.

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

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.

Returns: 3

SSR(1, 1), SSR(4, 1) and SSR(9, 1) are integers.

Returns: 5

The valid pairs are (1, 1), (1, 4), (2, 2), (2, 8) and (3, 3).

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++;

}