On XY plane, you are given upper left corner coordinates of two equal sized rect

Question

On XY plane, you are given upper left corner coordinates of two equal sized rectangles. Find the intersection percentage of the rectangles.

Question: Write a program that is going to read two rectangle information and find the intersection percentage.

Input specification
You will be first given the width (w) and height (h) of the rectangles. Then upper left coordinates (x, y) of the two rectangles are given in the following two lines where 0 <= (x, y) <= 10000 and 1 <= (w, h) <= 100.

Output specification
Show a floating point number with 3 decimal places that is the intersection percentage to the second rectangle.

Input
4 3
1 4
3 5

Output
33.333

note:use in c++ while, for , if loops ,do not use return or calculation

Explanation / Answer

Hello,

Below is the program for your question -

int main(int argc, const char * argv[])

{

int width,height,xA1,yA1,xB1,yB1;

cin>>width>>height; // input width and height

cin>>xA1>>yA1; //input 1st rectangle coordinates

cin>>xB1>>yB1; //input 2nd rectangle coordinates

// bottom most coordinates of 1st rectangle

int xA2=xA1-width;

int yA2=yA1-height;

// bottom most coordinates of 2nd rectangle  

int xB2=xB1-width;

int yB2=yB1-height;

int area=(width*height);

int widthIntersection;

int heightIntersection;

int areaIntersection=0;

   float intersectionPercentage=0.0

if(yA1<yB2 || yB1<yA2 || xA1<xB2 || xB1<xA2){ // to check if rectangles will intersect.

fixed << setprecision(3)<<intersectionPercentage;

}

else{

widthIntersection=abs(max(xA2,xB2)-min(xA1,xB1));

heightIntersection=abs(max(yA2,yB2)-min(yA1,yB1));

areaIntersection=widthIntersection*heightIntersection;

int unionArea=2*area-2*areaIntersection-areaIntersection;

intersectionPercentage=((float)areaIntersection/(float)unionArea)*100;

cout<< fixed << setprecision(3)<<intersectionPercentage;

}

}

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