The darker shadow of an eclipse is called the penumbra. True False

What are the raw materials for the process of photosynthesis

Ray Of Light [21]

The raw materials of photosynthesis, water and carbon dioxide, enter the cells of the leaf, and the products of photosynthesis, sugar and oxygen, leave the leaf.

7 0

2 years ago

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Write a public interface named Test with the following behavior:

enyata [817]

Answer:

The code to this question can be given as:

Code:

public interface Test //define interface

{

public abstract Duration getDuration(); //define method

getDuration.

public abstract Result check(int a);

//define method

check .

public abstract double getScore();

//define method getScore.

}

Explanation:

In the above code, we define an interface that is "Test" inside an interface, we define three methods that can be defined as:

First, we define a method that is "getDuration" that method returns Duration as an object.
Second, we define a method that is "check" this method accepts an integer parameter that is a and return Result.
The third method is "getScore" this method will return a double value.

8 0

3 years ago

Structure for forloop?

solong [7]

Answer:

for ( initialization; condition;increment)

{

code goes here;

}

in python:

for i in list/range:

code with proper indentation

By initialization above we mean, like int i=0; etc.

By condition like i<10;

and by increment it means like i++, ++i or i+=1; etc

And in python, i can be an integer value if the range is mentioned, and it can be an item of a list if the list is used. We can also use an array, string and various other data structures in python. like we can have characters in a string and so on.

Explanation:

Please check the answer section.

8 0

3 years ago

How do I mark someone brainiest

svet-max [94.6K]

Answer:

when someone answers your questio correctly, you will see a red crown. Click on that and mark them the most brainiest

Explanation:

3 0

2 years ago

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a cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits such that the giv

Leokris [45]

Using the knowledge in computational language in C++ it is possible to write a code that cryptarithm is a mathematical puzzle where the goal is to find the correspondence between letters and digits

<h3>Writting the code:</h3>

#include <bits/stdc++.h>

using namespace std;

// chracter to digit mapping, and the inverse

// (if you want better performance: use array instead of unordered_map)

unordered_map<char, int> c2i;

unordered_map<int, char> i2c;

int ans = 0;

// limit: length of result

int limit = 0;

// digit: index of digit in a word, widx: index of a word in word list, sum: summation of all word[digit]

bool helper(vector<string>& words, string& result, int digit, int widx, int sum) {

if (digit == limit) {

ans += (sum == 0);

return sum == 0;

}

// if summation at digit position complete, validate it with result[digit].

if (widx == words.size()) {

if (c2i.count(result[digit]) == 0 && i2c.count(sum%10) == 0) {

if (sum%10 == 0 && digit+1 == limit) // Avoid leading zero in result

return false;

c2i[result[digit]] = sum % 10;

i2c[sum%10] = result[digit];

bool tmp = helper(words, result, digit+1, 0, sum/10);

c2i.erase(result[digit]);

i2c.erase(sum%10);

ans += tmp;

return tmp;

} else if (c2i.count(result[digit]) && c2i[result[digit]] == sum % 10){

if (digit + 1 == limit && 0 == c2i[result[digit]]) {

return false;

}

return helper(words, result, digit+1, 0, sum/10);

} else {

return false;

}

// if word[widx] length less than digit, ignore and go to next word

if (digit >= words[widx].length()) {

return helper(words, result, digit, widx+1, sum);

}

// if word[widx][digit] already mapped to a value

if (c2i.count(words[widx][digit])) {

if (digit+1 == words[widx].length() && words[widx].length() > 1 && c2i[words[widx][digit]] == 0)

return false;

return helper(words, result, digit, widx+1, sum+c2i[words[widx][digit]]);

}

// if word[widx][digit] not mapped to a value yet

for (int i = 0; i < 10; i++) {

if (digit+1 == words[widx].length() && i == 0 && words[widx].length() > 1) continue;

if (i2c.count(i)) continue;

c2i[words[widx][digit]] = i;

i2c[i] = words[widx][digit];

bool tmp = helper(words, result, digit, widx+1, sum+i);

c2i.erase(words[widx][digit]);

i2c.erase(i);

}

return false;

}

void isSolvable(vector<string>& words, string result) {

limit = result.length();

for (auto &w: words)

if (w.length() > limit)

return;

for (auto&w:words)

reverse(w.begin(), w.end());

reverse(result.begin(), result.end());

int aa = helper(words, result, 0, 0, 0);

}

int main()

{

ans = 0;

vector<string> words={"GREEN" , "BLUE"} ;

string result = "BLACK";

isSolvable(words, result);

cout << ans << "\n";

return 0;

}

See more about C++ code at brainly.com/question/19705654

#SPJ1

3 0

2 years ago