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PtichkaEL [24]
3 years ago
5

write a program that reads in the length and the width of a rectangular yard . your program should compute the time required ( i

n minutes ) to cut the grass at the rate of 2.3 square meters a second
Computers and Technology
1 answer:
sveticcg [70]3 years ago
3 0
Here's on C:
#include<stdlib.h> 
<span>#include<stdio.h> </span>
<span>#include<conio.h> </span>
<span>long YardArea=0,HouseArea=0; </span>
<span>int YardLength=0,YardWidth=0,HouseLength=0,HouseWidth=0; </span>
<span>long GroundArea=0; </span>
<span>int TimeReqierd=0; </span>
<span>int GrassCutRate=2; </span>
<span>void main() </span>
<span>{ </span>
<span>char inp[256]; </span>
<span>clrscr(); </span>
<span>printf("your text text\n");  // put your greetings</span>
<span>printf("Input the leght of the yard\n"); </span>
<span>gets(inp); </span>
<span>YardLength= atoi(inp); </span>
<span>printf("input the width of the yard\n"); </span>
<span>         gets(inp); </span>
<span>YardWidth=atoi(inp); </span>
<span>    printf("input the length of the house\n"); </span>
<span>         gets(inp); </span>
<span>HouseLength=atoi(inp); </span>
   printf("input <span>the width of the house\n"); </span>
<span>         gets(inp); </span>
<span>HouseWidth=atoi(inp); </span>
<span>HouseArea=HouseWidth *HouseLength ; </span>
<span>YardArea=YardWidth* YardLength; </span>
<span>GroundArea=YardArea-HouseArea; </span>
<span>   TimeReqierd=GroundArea/GrassCutRate; </span>
<span>printf("%s%d","Grass Area is ",GroundArea); </span>
<span>printf("%s%d seconds"," \n time required to cut it out " ,TimeReqierd); </span>
<span>getchar(); </span>
<span>} </span>
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