Write a program which displays counting less than 100?
408935
Hey,\nthe code in JAVA would look like this ::\n\nclass test {\npublic static void main(String a[]) {\nfor(int i=1;i<100;i++) {\nSystem.out.print("&#92;t"+i+"&#92;n");\n}\n}\n}
73238
Try, http://www.simplyscripts.com/
368941
I can help you with this program in C (not C++).\n\n#include <stdio.h>\n\nint a;\n\nvoid main(void)\n{\n   for(a = 0; a <= 1000; a++)\n   {\n       printf("%d&#92;n", a);\n   }\n}\n\nThe above program lists numbers starting from 0 to 1000.\nYou can modify the program to add up all those numbers if that is what you want.\n\n#include <stdio.h>\n\nint a, sum = 0;\n\nvoid main(void)\n{\n   for(a = 0; a <= 1000; a++)\n   {\n      printf("%d&#92;n", a);\n      sum = sum + a;\n   }\n   printf("Sum = %d", sum);\n}\n\nThis program prints and also finds the sum to 1000 numbers starting from 0!
318310
take the percent number and put it over 100. then simplify
985771
You can generate this in one line of Perl, Unix scripting, or a small C/C++ program.  Simply write one number per line from your program counter.  You may have to get tricky and use two long ints, but it's not hard.\n\n Realize that your document may be very large, and for most purposes, you don't need a document, just use the variable(s) in your final program.
297664
Wrap-around, modular arithmetic.  If the sum exceeds 100, subtract 100 so the result is still 2-digit.  And when decoding, if the result goes negative, add 100.
469353
divide the numbers (not the percentages) by 100
158397
C solution:\n\nint nr[20];\nint valid=1,i,j;\nint main(int argc, char **argv){\nfor (i=0;(i<20)&&(valid);i++){\n scanf("%d",&nr[i]);\n if (nr[i]<10)||(nr[i]>100){\n  printf("Invalid number");\n  valid=0;\n }\n for (j=0;j<i;j++)\n  if (nr[j]==nr[i]){\n    printf("Duplicate number");\n    valid=0;\n  }\n}\n}
253284
#include <stdio.h>\n\nint main(void) {\n\nint i, j, count; //C Sucks, wait till you take java\n\nfor(i=1; i<=1000; i++){&#92;&#92;goes through all numbers 1 to 1000\n    count=0;\n    for(j=2; j<i; j++)&#92;&#92;goes through all numbers 2 to i\n    {\n     if(i%j==0)&#92;&#92;checks to see if it i is evenly div. by i\n        count++;\n    }//nested for\n    if (count ==0)\n        printf("%d&#92;n", i);\n}//for\n\nreturn 0;\n}//main\n\n\n//Hope that helps!
656675
include <stdio.h>\n\nvoid main(){\n   float no;\n   scanf("%f",&no);\n   if(!( ((int)no)%7!=0 || ((int)no)%2==0 )) \n      printf("No is divisible by 7 and not divisible by 2");\n}\n\nsalam
551157
Alright. Here's the deal. \n\nSince this is your first course, here are some tricks.\n\nIf the program will have an input of name, month,and day the hurricane, then you will need 4 variables. A string and 3 integers.\n\nHow will you use them? How many times will you repeat the loop?\n\nIf you do not know how many times you want it to be repeated, then  the wisest choice is a do/while. Otherwise, let the user tell you how many times it should be done.\n\nYou will make those 4 variables. They should be arrays, make sure they are big enough.\n\nit will be something as simple as this:\n\nlet the user tell you how many times he want it to be repeated:\n\nint timesToBeRepeated = 0;\ncin >> timesToBeRepeated;\n\nfor ( int i = 0; i < timesToBeRepeated; i++ )\n{\n\n      cout <<"Enter the Hurricane name: &#92;n";\n      cin >> hurricaneName[i];\n         .\n         .\n         .\n}\n\nthen, to print it out,  just do something like this:\n\ncout <<"Hurricane Name"<<"      <<"Day"<<..<<endl;\n\nthen:\n\nfor ( i = 0; i < timesToBeRepeated; i++ )\n  cout << hurricaneName[i] <<"   "<<....<<endl;\n\nWork out the rest, but that is pretty much it., Any doubt, contact me.
1078011
Conditional statements only include those using if, if else, if else if, and switch, so while, do while, and for loops should be acceptable.  The answer is then either:\n\nint x = 1;\ndo{\n  printf(“&#92;n”, x);\n  x++;\n}while (x < 100);\ndo{\n  printf(“&#92;n”, x);\n  x--;\n}while (x > 0);\n\nOR:\nfor (int x = 1; x < 100; x++){\n  printf(“&#92;n”, x);\n}\nfor (x = 100; x < 0; x--){\n  printf(“&#92;n”, x);\n}
1158354
This does not require hiring a professional, it's a rather simple & straightforward program. \n\nRaven put you on the right track, use the modulo (%) operator to break out the thousands from the hundreds, the hundreds from the tens, etc.  Define an array or map of strings with the words for each number up to twenty, then thirty, forty, fifty.. to ninety, then hundred, thousand.  Break the number apart, and get the appropriate words from the array & string them together.
166754
Just have a variable called max and min\nfor the first number then enter just set that to max and min\nthen for each additional number\nif num > max then max=num\nif num < min and num <> -99 then min=num
982889
Examples (in Visual Basic):\n\nx = 1\nDo While x < 10\n    MsgBox "Less Than 10!"\n    x = x + 1\nLoop\n\nFor x = 1 to 9\n  MsgBox "Less Than 10!"\nNext x\n\nBoth will give the same result, but a different way of doing it.  You can get each to do different things or the same thing, depending on how you write them.  Do While Loops do something WHILE something is true (x<10) and For Next Loops do something for a fixed amount of time.
1024904
Thats a recursive program for you.\nAnd I put an arbitrary limit of 9 there. You can change that for yourself.\n\n#include <stdio.h>\n\nvoid triangle(char ch, int count) {\n        int temp=0;\n        if (count==1) {\n                printf("%c&#92;n", ch);\n                return;\n        }\n        triangle(ch, count-1);\n        for(;temp<count;temp++) {\n                printf("%c", ch);\n        }\n        printf("&#92;n");\n}\n\nint main() {\n        char ch='*';\n        int count=0;\n        while((count<=0)||(count>=10)) {\n                printf("Enter the size of triangle (less than 10)&#92;n");\n                scanf("%d", &count);\n        }\n        triangle(ch, count);\n}
926416
your getting a little anxious in your coding:\ntry changing the section below:\n\ncout << "Enter a number Negative, Positve, or Zero.";\ncout << '&#92;n';\ncin >> num;\n\nif (num > 0)\n{\n    pos(num);\n}\nelse if (num < 0)\n{\n  neg(num);\n}\nelse\n{\n  zero(num);\n}