Integers program C++?
959017
I see two issues, in addition to the division problem pointed out above.  First, you should initialize your variables.  In particular, set sum=0 at the beginning of your code. Do the same for counter.\n\nNext, you have logic issues.  For min and max, you have to initialize these variables, but just setting them to 0 may not work.  If the user enters a series of negative values, you may falsely report 0 as a max. So, the classic solution is to initialize min and max to the first number given by the user.  Of course, this means you'll have to prompt the user for input outside the loop.\n\nBut actually, this helps to solve the other problem.  In particular, when the user enters 0, you still increment your counter. It would be best to redesign your code a little so that the loop condition guarantees that the most recent user value is not zero. You can do this by changing the type of loop that you use, and by asking for the first value outside the loop. \n\nI haven't quite given you all the details, but that's because I want you to think about it yourself.  It'll be a valuable exercise.
440005
Pseudocode, I'm not up on my Java syntax:\n\nint sum=0;\nfor(int i=0;i<array.size;i++)\n  sum+=array[i];
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.
1065652
void main()\n{\ncout<<"no"<<"/t"<<"id.no"<<"/t"<<"name"<<"/t"<<"sex"<<"/t"<<"department";\ncout<<"1 001/98 sarafcomputerscience&#92;n";\ncout<<"your information";\ncout<<"your information";\ncout<<"your information";\ncout<<"your information";\n\n}\nif u want to get input at runtime try to input thr cin>> statement and get that output by cout statement
345718
int main (void)\n{\nint a,rem\ncout<<"enter the number";\n<<cin a;\nrem=a/a;\nif(rem>0)\ncout<<"its integer";\nelse\ncout<<"not a integer";\nreturn 0;\n}
662225
zero is an integer.\nAn integer is any whole number (positive or negative), including zero.
457026
Create a method called numDigits(long). Then you can use code something like this:\n\nint num;\ncin >> num;\n\nint numDigs = numDigits( num );\nint digits[ numDigs ];\nint temp = num;\nfor ( int i = numDigs - 1; i >= 0; i-- )\n{\n digits[ i ] = temp % 10;\n temp = (int)( temp / 10 );\n}\n\nThe digits array holds each of the digits in the number.\nThis is the code I used for numDigits():\n\nint numDigits( long num )\n{\nint digits;\nfor ( digits = 0; num > 0; digits++ )\n{\nnum /= 10;\n}\n\nreturn digits;\n}
966120
try this stuff... pls understand the logic...\n\n#include<stdio.h>\n\nint main()\n{\nint no,rem;\nscanf("%d",&no);\nwhile (no)\n{\nrem = no % 10;\nprintf("%d",rem);\nno = no / 10;\n}\n}
129073
The code below will work.\n\nint main()\n{\nint aNum,i;\ncout<<"enter a number";\ncin>> aNum;\nfor(i=0;i<aNum;i++)\n{\nfor(j=0;j<=i;j++)//nested loop\ncout<<"*";// end of nested loop\ncout<<'&#92;n';\n}// end of outer loop\nreturn 0;\n}
557543
#include<iostream>\nusing namespace std;\nint main ()\n\n\n    {\nint number;\nint answer;\n    cout<<"Please enter number: ";\n    cin>>number;\n    \n    cout<<number<<"! = ";\n    answer = number;\n    while (number>1)\n\n\n        {\n        answer=answer*(number-1);\n        number--; \n    } \n\n    cout<<answer; \n    return 0; \n}
878923
Make a program on your calculator.\nView the snippet in the link for code.
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
1084101
You must cast legs to an integer using parseInt\n\nlegs = parseInt(legs);
515624
You can start off by declaring a new integer variable "low", initialize it to 0, and then after any input compare the new value to the value in low and if the new value is lower then set low equal to the new value.\nThe following sites should also be of use.
288377
PASCAL!!!its out dated, i think it is possible to get it on C/C++
742648
#include<stdio.h>\n#include<conio.h>\nvoid main()\n{\nint n=5,i,j;\nclrscr();\nfor(i=1;i<=2*n-2;i++)\n{\nprintf(" ");\n}\nprintf("1&#92;n");\nfor(i=2;i<=n-1;i++)\n{\nfor(j = 1;j<=(n-i)*2;j++)\nprintf(" ");\nfor(j = i;j<=2*i-1;j++)\n{\nprintf("%d ",j);\n}\nfor(j = 2*i-2;j>=i;j--)\n{\nprintf("%d ",j);\n}\nprintf("&#92;n");\n}\nfor(i = n;i<=2*n-1;i++)\nprintf("%d ",i);\nfor(i = 2*n-2;i>=n;i--)\nprintf("%d ",i);\ngetch();\n}
560969
23\nif intead of '=' there is +