有些眼镜的容量等于1升。眼镜的存放方法如下：
1
2 3
4 5 6
7 8 9 10您可以将水倒入唯一的顶部玻璃杯中。如果您在第一个玻璃杯中放了超过1升的水，则第二个和第三个玻璃杯中的水会溢出并均匀地填充。玻璃5将从第二玻璃和第三玻璃中取水，依此类推。
如果您有X升水，然后将其放在顶玻璃杯中，那么第i排第j杯玻璃中会容纳多少水?
例子。如果您将2升放在顶部。
1 – 1升
2 – 1/2升
3 – 1/2升
2公升水
该方法类似于Pascal三角形的方法2。如果我们仔细研究这个问题，这个问题可以归结为帕斯卡的三角形。
1 ---------------- 1
2 3 ---------------- 2
4 5 6 ------------ 3
7 8 9 10 --------- 4每个玻璃杯都有助于两个玻璃杯。最初，我们将所有水倒入第一杯中。然后，将1升(或少于1升)的水倒入其中，并将剩下的水倒入两杯中。我们对两副眼镜和所有其他眼镜遵循相同的过程，直到第i行为止。到第i行为止，将有i *(i + 1)/ 2杯。
C++
// Program to find the amount of water in j-th glass
// of i-th row
#include
#include
#include
// Returns the amount of water in jth glass of ith row
float findWater(int i, int j, float X)
{
// A row number i has maximum i columns. So input
// column number must be less than i
if (j > i)
{
printf("Incorrect Inputn");
exit(0);
}
// There will be i*(i+1)/2 glasses till ith row
// (including ith row)
float glass[i * (i + 1) / 2];
// Initialize all glasses as empty
memset(glass, 0, sizeof(glass));
// Put all water in first glass
int index = 0;
glass[index] = X;
// Now let the water flow to the downward glasses
// till the row number is less than or/ equal to i (given row)
// correction : X can be zero for side glasses as they have lower rate to fill
for (int row = 1; row <= i ; ++row)
{
// Fill glasses in a given row. Number of
// columns in a row is equal to row number
for (int col = 1; col <= row; ++col, ++index)
{
// Get the water from current glass
X = glass[index];
// Keep the amount less than or equal to
// capacity in current glass
glass[index] = (X >= 1.0f) ? 1.0f : X;
// Get the remaining amount
X = (X >= 1.0f) ? (X - 1) : 0.0f;
// Distribute the remaining amount to
// the down two glasses
glass[index + row] += X / 2;
glass[index + row + 1] += X / 2;
}
}
// The index of jth glass in ith row will
// be i*(i-1)/2 + j - 1
return glass[i*(i-1)/2 + j - 1];
}
// Driver program to test above function
int main()
{
int i = 2, j = 2;
float X = 2.0; // Total amount of water
printf("Amount of water in jth glass of ith row is: %f",
findWater(i, j, X));
return 0;
} Java
// Program to find the amount
/// of water in j-th glass
// of i-th row
import java.lang.*;
class GFG
{
// Returns the amount of water
// in jth glass of ith row
static float findWater(int i, int j,
float X)
{
// A row number i has maximum i
// columns. So input column
// number must be less than i
if (j > i)
{
System.out.println("Incorrect Input");
System.exit(0);
}
// There will be i*(i+1)/2 glasses
// till ith row (including ith row)
int ll = Math.round((i * (i + 1) ));
float[] glass = new float[ll + 2];
// Put all water in first glass
int index = 0;
glass[index] = X;
// Now let the water flow to the
// downward glasses till the row
// number is less than or/ equal
// to i (given row)
// correction : X can be zero for side
// glasses as they have lower rate to fill
for (int row = 1; row <= i ; ++row)
{
// Fill glasses in a given row. Number of
// columns in a row is equal to row number
for (int col = 1;
col <= row; ++col, ++index)
{
// Get the water from current glass
X = glass[index];
// Keep the amount less than or
// equal to capacity in current glass
glass[index] = (X >= 1.0f) ? 1.0f : X;
// Get the remaining amount
X = (X >= 1.0f) ? (X - 1) : 0.0f;
// Distribute the remaining amount
// to the down two glasses
glass[index + row] += X / 2;
glass[index + row + 1] += X / 2;
}
}
// The index of jth glass in ith
// row will be i*(i-1)/2 + j - 1
return glass[(int)(i * (i - 1) /
2 + j - 1)];
}
// Driver Code
public static void main(String[] args)
{
int i = 2, j = 2;
float X = 2.0f; // Total amount of water
System.out.println("Amount of water in jth " +
"glass of ith row is: " +
findWater(i, j, X));
}
}
// This code is contributed by mitsPython3
# Program to find the amount
# of water in j-th glass of
# i-th row
# Returns the amount of water
# in jth glass of ith row
def findWater(i, j, X):
# A row number i has maximum
# i columns. So input column
# number must be less than i
if (j > i):
print("Incorrect Input");
return;
# There will be i*(i+1)/2
# glasses till ith row
# (including ith row)
# and Initialize all glasses
# as empty
glass = [0]*int(i *(i + 1) / 2);
# Put all water
# in first glass
index = 0;
glass[index] = X;
# Now let the water flow to
# the downward glasses till
# the row number is less
# than or/ equal to i (given
# row) correction : X can be
# zero for side glasses as
# they have lower rate to fill
for row in range(1,i):
# Fill glasses in a given
# row. Number of columns
# in a row is equal to row number
for col in range(1,row+1):
# Get the water
# from current glass
X = glass[index];
# Keep the amount less
# than or equal to
# capacity in current glass
glass[index] = 1.0 if (X >= 1.0) else X;
# Get the remaining amount
X = (X - 1) if (X >= 1.0) else 0.0;
# Distribute the remaining
# amount to the down two glasses
glass[index + row] += (X / 2);
glass[index + row + 1] += (X / 2);
index+=1;
# The index of jth glass
# in ith row will
# be i*(i-1)/2 + j - 1
return glass[int(i * (i - 1) /2 + j - 1)];
# Driver Code
if __name__ == "__main__":
i = 2;
j = 2;
X = 2.0;
# Total amount of water
res=repr(findWater(i, j, X));
print("Amount of water in jth glass of ith row is:",res.ljust(8,'0'));
# This Code is contributed by mitsC#
// Program to find the amount
// of water in j-th glass
// of i-th row
using System;
class GFG
{
// Returns the amount of water
// in jth glass of ith row
static float findWater(int i, int j,
float X)
{
// A row number i has maximum i
// columns. So input column
// number must be less than i
if (j > i)
{
Console.WriteLine("Incorrect Input");
Environment.Exit(0);
}
// There will be i*(i+1)/2 glasses
// till ith row (including ith row)
int ll = (int)Math.Round((double)(i * (i + 1)));
float[] glass = new float[ll + 2];
// Put all water in first glass
int index = 0;
glass[index] = X;
// Now let the water flow to the
// downward glasses till the row
// number is less than or/ equal
// to i (given row)
// correction : X can be zero
// for side glasses as they have
// lower rate to fill
for (int row = 1; row <= i ; ++row)
{
// Fill glasses in a given row.
// Number of columns in a row
// is equal to row number
for (int col = 1;
col <= row; ++col, ++index)
{
// Get the water from current glass
X = glass[index];
// Keep the amount less than
// or equal to capacity in
// current glass
glass[index] = (X >= 1.0f) ?
1.0f : X;
// Get the remaining amount
X = (X >= 1.0f) ? (X - 1) : 0.0f;
// Distribute the remaining amount
// to the down two glasses
glass[index + row] += X / 2;
glass[index + row + 1] += X / 2;
}
}
// The index of jth glass in ith
// row will be i*(i-1)/2 + j - 1
return glass[(int)(i * (i - 1) /
2 + j - 1)];
}
// Driver Code
static void Main()
{
int i = 2, j = 2;
float X = 2.0f; // Total amount of water
Console.WriteLine("Amount of water in jth " +
"glass of ith row is: " +
findWater(i, j, X));
}
}
// This code is contributed by mitsPHP
$i)
{
echo "Incorrect Input\n";
return;
}
// There will be i*(i+1)/2
// glasses till ith row
// (including ith row)
// and Initialize all glasses
// as empty
$glass = array_fill(0, (int)($i *
($i + 1) / 2), 0);
// Put all water
// in first glass
$index = 0;
$glass[$index] = $X;
// Now let the water flow to
// the downward glasses till
// the row number is less
// than or/ equal to i (given
// row) correction : X can be
// zero for side glasses as
// they have lower rate to fill
for ($row = 1; $row < $i ; ++$row)
{
// Fill glasses in a given
// row. Number of columns
// in a row is equal to row number
for ($col = 1;
$col <= $row; ++$col, ++$index)
{
// Get the water
// from current glass
$X = $glass[$index];
// Keep the amount less
// than or equal to
// capacity in current glass
$glass[$index] = ($X >= 1.0) ?
1.0 : $X;
// Get the remaining amount
$X = ($X >= 1.0) ?
($X - 1) : 0.0;
// Distribute the remaining
// amount to the down two glasses
$glass[$index + $row] += (double)($X / 2);
$glass[$index + $row + 1] += (double)($X / 2);
}
}
// The index of jth glass
// in ith row will
// be i*(i-1)/2 + j - 1
return $glass[(int)($i * ($i - 1) /
2 + $j - 1)];
}
// Driver Code
$i = 2;
$j = 2;
$X = 2.0; // Total amount of water
echo "Amount of water in jth " ,
"glass of ith row is: ".
str_pad(findWater($i, $j,
$X), 8, '0');
// This Code is contributed by mits
?>Javascript
C++
// CPP program for above approach
#include
using namespace std;
// Program to find the amount
// of water in j-th glass
// of i-th row
void findWater(float k, int i, int j)
{
// stores how much amount of water
// present in every glass
float dp[i+1][2*i + 1];
bool vis[i+1][2*i + 1];
// initialize dp and visited arrays
for(int n=0;n>q;
dp[0][i] = k;
// we take the center of the first row for
// the location of the first glass
q.push({0,i});
vis[0][i] = true;
// this while loop runs unless the queue is not empty
while(!q.empty())
{
// First we remove the pair from the queue
pairtemp = q.front();
q.pop();
int n = temp.first;
int m = temp.second;
// as we know we have to calculate the
// amount of water in jth glass
// of ith row . so first we check if we find solutions
// for the the every glass of i'th row.
// so, if we have calculated the result
// then we break the loop
// and return our answer
if(i == n)
break;
float x = dp[n][m];
if(float((x-1.0)/2.0) < 0)
{
dp[n+1][m-1] += 0;
dp[n+1][m+1] += 0;
}
else
{
dp[n+1][m-1] += float((x-1.0)/2.0);
dp[n+1][m+1] += float((x-1.0)/2.0);
}
if(vis[n+1][m-1]==false)
{
q.push({n+1,m-1});
vis[n+1][m-1] = true;
}
if(vis[n+1][m+1]==false)
{
q.push({n+1,m+1});
vis[n+1][m+1] = true;
}
}
if(dp[i-1][2*j-1]>1)
dp[i-1][2*j-1] = 1.0;
cout<<"Amount of water in jth glass of ith row is: ";
// print the result for jth glass of ith row
cout<water in litres
float k;
cin>>k;
//i->rows j->cols
int i,j;
cin>>i>>j;
// Function Call
findWater(k,i,j);
return 0;
}
// This code is contibuted by Sagar Jangra and Naresh Saharan Java
// Program to find the amount
/// of water in j-th glass
// of i-th row
import java.io.*;
import java.util.*;
// class Triplet which stores curr row
// curr col and curr rem-water
// of every glass
class Triplet
{
int row;
int col;
double rem_water;
Triplet(int row,int col,double rem_water)
{
this.row=row;this.col=col;this.rem_water=rem_water;
}
}
class GFG
{
// Returns the amount of water
// in jth glass of ith row
public static double findWater(int i,int j,int totalWater)
{
// stores how much amount of water present in every glass
double dp[][] = new double[i+1][2*i+1];
// store Triplet i.e curr-row , curr-col, rem-water
Queue queue = new LinkedList<>();
// we take the center of the first row for
// the location of the first glass
queue.add(new Triplet(0,i,totalWater));
// this while loop runs unless the queue is not empty
while(!queue.isEmpty())
{
// First we remove the Triplet from the queue
Triplet curr = queue.remove();
// as we know we have to calculate the
// amount of water in jth glass
// of ith row . so first we check if we find solutions
// for the the every glass of i'th row.
// so, if we have calculated the result
// then we break the loop
// and return our answer
if(curr.row == i)
break;
// As we know maximum capacity of every glass
// is 1 unit. so first we check the capacity
// of curr glass is full or not.
if(dp[curr.row][curr.col] != 1.0)
{
// calculate the remaining capacity of water for curr glass
double rem_water = 1-dp[curr.row][curr.col];
// if the remaining capacity of water for curr glass
// is greater than then the remaining capacity of tatal
// water then we put all remaining water into curr glass
if(rem_water > curr.rem_water)
{
dp[curr.row][curr.col] += curr.rem_water;
curr.rem_water = 0;
}
else
{
dp[curr.row][curr.col] += rem_water;
curr.rem_water -= rem_water;
}
}
// if remaining capacity of tatal water is not equal to
// zero then we add left and right glass of the next row
// and gives half capacity of tatal water to both the
// glass
if(curr.rem_water != 0)
{
queue.add(new Triplet(curr.row + 1,curr.col -
1,curr.rem_water/2.0));
queue.add(new Triplet(curr.row + 1,curr.col +
1,curr.rem_water/2.0));
}
}
// return the result for jth glass of ith row
return dp[i-1][2*j-1];
}
// Driver Code
public static void main (String[] args)
{
int i = 2, j = 2;
int totalWater = 2; // Total amount of water
System.out.print("Amount of water in jth glass of ith row is:");
System.out.format("%.6f", findWater(i, j, totalWater));
}
}
// this code is contibuted by Naresh Saharan and Sagar Jangra 输出：
Amount of water in jth glass of ith row is: 0.500000方法2(使用BFS遍历)
我们将讨论解决此问题的另一种方法。首先，我们在队列中添加一个Triplet(row，col，rem-water)，它指示第一个元素的起始值并填充1升水。然后我们简单地应用bfs即，我们将left(row + 1，col-1，rem-water)Triplet和right(row + 1，col + 1，rem-water)Triplet加到队列中，剩下的一半留在第一个三胞胎，另一半进入下一个三胞胎。
以下是此解决方案的实现。
C++
// CPP program for above approach
#include
using namespace std;
// Program to find the amount
// of water in j-th glass
// of i-th row
void findWater(float k, int i, int j)
{
// stores how much amount of water
// present in every glass
float dp[i+1][2*i + 1];
bool vis[i+1][2*i + 1];
// initialize dp and visited arrays
for(int n=0;n>q;
dp[0][i] = k;
// we take the center of the first row for
// the location of the first glass
q.push({0,i});
vis[0][i] = true;
// this while loop runs unless the queue is not empty
while(!q.empty())
{
// First we remove the pair from the queue
pairtemp = q.front();
q.pop();
int n = temp.first;
int m = temp.second;
// as we know we have to calculate the
// amount of water in jth glass
// of ith row . so first we check if we find solutions
// for the the every glass of i'th row.
// so, if we have calculated the result
// then we break the loop
// and return our answer
if(i == n)
break;
float x = dp[n][m];
if(float((x-1.0)/2.0) < 0)
{
dp[n+1][m-1] += 0;
dp[n+1][m+1] += 0;
}
else
{
dp[n+1][m-1] += float((x-1.0)/2.0);
dp[n+1][m+1] += float((x-1.0)/2.0);
}
if(vis[n+1][m-1]==false)
{
q.push({n+1,m-1});
vis[n+1][m-1] = true;
}
if(vis[n+1][m+1]==false)
{
q.push({n+1,m+1});
vis[n+1][m+1] = true;
}
}
if(dp[i-1][2*j-1]>1)
dp[i-1][2*j-1] = 1.0;
cout<<"Amount of water in jth glass of ith row is: ";
// print the result for jth glass of ith row
cout<water in litres
float k;
cin>>k;
//i->rows j->cols
int i,j;
cin>>i>>j;
// Function Call
findWater(k,i,j);
return 0;
}
// This code is contibuted by Sagar Jangra and Naresh Saharan
Java
// Program to find the amount
/// of water in j-th glass
// of i-th row
import java.io.*;
import java.util.*;
// class Triplet which stores curr row
// curr col and curr rem-water
// of every glass
class Triplet
{
int row;
int col;
double rem_water;
Triplet(int row,int col,double rem_water)
{
this.row=row;this.col=col;this.rem_water=rem_water;
}
}
class GFG
{
// Returns the amount of water
// in jth glass of ith row
public static double findWater(int i,int j,int totalWater)
{
// stores how much amount of water present in every glass
double dp[][] = new double[i+1][2*i+1];
// store Triplet i.e curr-row , curr-col, rem-water
Queue queue = new LinkedList<>();
// we take the center of the first row for
// the location of the first glass
queue.add(new Triplet(0,i,totalWater));
// this while loop runs unless the queue is not empty
while(!queue.isEmpty())
{
// First we remove the Triplet from the queue
Triplet curr = queue.remove();
// as we know we have to calculate the
// amount of water in jth glass
// of ith row . so first we check if we find solutions
// for the the every glass of i'th row.
// so, if we have calculated the result
// then we break the loop
// and return our answer
if(curr.row == i)
break;
// As we know maximum capacity of every glass
// is 1 unit. so first we check the capacity
// of curr glass is full or not.
if(dp[curr.row][curr.col] != 1.0)
{
// calculate the remaining capacity of water for curr glass
double rem_water = 1-dp[curr.row][curr.col];
// if the remaining capacity of water for curr glass
// is greater than then the remaining capacity of tatal
// water then we put all remaining water into curr glass
if(rem_water > curr.rem_water)
{
dp[curr.row][curr.col] += curr.rem_water;
curr.rem_water = 0;
}
else
{
dp[curr.row][curr.col] += rem_water;
curr.rem_water -= rem_water;
}
}
// if remaining capacity of tatal water is not equal to
// zero then we add left and right glass of the next row
// and gives half capacity of tatal water to both the
// glass
if(curr.rem_water != 0)
{
queue.add(new Triplet(curr.row + 1,curr.col -
1,curr.rem_water/2.0));
queue.add(new Triplet(curr.row + 1,curr.col +
1,curr.rem_water/2.0));
}
}
// return the result for jth glass of ith row
return dp[i-1][2*j-1];
}
// Driver Code
public static void main (String[] args)
{
int i = 2, j = 2;
int totalWater = 2; // Total amount of water
System.out.print("Amount of water in jth glass of ith row is:");
System.out.format("%.6f", findWater(i, j, totalWater));
}
}
// this code is contibuted by Naresh Saharan and Sagar Jangra
输出
Amount of water in jth glass of ith row is:0.500000时间复杂度： O(2 ^ i)
空间复杂度： O(i ^ 2)