How can i put numbers in ascending order in c++?
604143
here is a program for 10 numbers in ascending order\n\n#include <iostream.h>\n#include <conio.h>\n\nvoid main()\n{\nclrscr();\nint array[10],t;\nfor(int x=0;x<10;x++)\n{\ncout << "Enter Integer No. " << x+1 << " : " << endl;\ncin>>array[x];\n}\nfor (x=0;x<10;x++)\n{\nfor(int y=0;y<9;y++)\n{\nif(array[y]>array[y+1])\n{\nt=array[y];\narray[y]=array[y+1];\narray[y+1]=t;\n}\n}\n}\ncout << "Array in ascending order is : ";\nfor (x=0;x<10;x++)\ncout << endl << array[x];\ngetch();\n}\n\nThis program takes in the ten integers array[x] to be stored in the single-dimensional array as a screen input from the user.\nIt then sorts out these ten integers into ascending order and prints them out using the 'cout' command.
138457
Assuming this is in C, this is an impossible problem given the main program you provided, because you're passing in the parameters by value rather than by reference.  So any changes you make to those values inside the function will disappear once you return to the main context.\n\nIf you're allowed to change the main program to use pointers, you could do this:\n\nvoid sort2 (int *a, int *b) {\n  int tmp;\n\n  if (*a > *b) {\n   tmp = *a;\n   *a = *b;\n   *b = tmp;\n  }\n}\n\nGiven your main program, you'd call it like this:\n  sort2(&u, &v);\n\nBut it may be that this assignment was given to you specifically to demonstrate that you can't change the value of variables that you pass in by value).
256646
I'll put a bit of more information than the previous two answers.. or it'll seem like I'm answering for 2 points ;-)\n\nA for loop in C++ basically has the syntax:\n\nfor([initializations];[condition check];[updates])\n{\n  //statement1;\n  //statement2;\n}\n\nInitializations: Initialize counters, iterators.. any data to be set or reset before the loop starts executing. Multiple initialization statements can be provided in a comma-separated-list.\n\nCondition Check: While the condition is true, keep iterating through the loop. Dont make this a list.\n\nUpdates: Update iterators, indices, counters etc. which need to be updated with every iteration of the loop. Again this can be a comma-separated-list.
9961
0,1,1,2,3\n\nto get the next number add the previous 2 numbers
274854
The difficult thing here is reading and writing the numbers. They must be transformed from ASCII and then the result transformed back to ASCII.I found a program that calculates the sum of 2 numbers.I hope it helps.
458677
There's a very explicit algorithm which explains\nhow to handle number, operators (including left\nor right precedence), functions, etc. on the \nWikipedia site for RPN (see reference #1).\n\nMostly, it's very straight-forward.  If it's an operand\n(ie. a number), then push it on the stack.  If it's an\noperator, then pop off as many operands are needed and\nevaluate.\n\nBut do look at the full algorithm for the details.\n\nOh yeah, for a full-blown implementation, see the source\ncode to the Unix "dc" (desk calculator) program.  The\nSolaris source is at reference #2.
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.
784795
numbering pages beginning from 1 in ascending order
241489
Try any of this code\nDim i as integer=Convert.ToInt32("your number as string")\nor\nDim i as integer=Val("your number as string")\nor\nDim i as integer=CInt("your number as string")
943372
u have to ponit it or use\n\nthese are the 2 way's:\n\n#include <iostream>\nusing namespcae std;\nint main()\n{\nint num[8];\nint index;\n\nnum[0] = 42;\nnum[1] = 2001;\nnum[2] = 7;\nnum[3] = 180;\nnum[4] = 99;\nnum[5] = 1993;\n\nfor(index = 0; index <= 5; index++)\n{\nstd::cout<< index[num] << endl;\n}\n system("pause");\nreturn 0;\n}\n\nor\n\n#include <iostream>\n#define MAX 20\nusing namespcae std;\n\nint main()\n{\n\nint a[MAX] = {23,43,4,5,34,343,12,14};\nint b;\n\ncout.setf(ios::right);\n\nfor(b = 0; b < MAX; b++){\n\ncout << endl << endl;\n\ncout << "The Elemnt in ";\n\ncout.width(2);\n\ncout << (b +1) << " is ";\n\ncout.width(2);\n\nstd::cout << b[a];\n}\nreturn 0;\n}
791300
Use the FOIL method; First, Outer, Inner, Last.\n\nFirst, n * n\n\nOuter, n * 5\n\nInner, n * 7\n\nLast 7 * 5.\n\nThen add them all together.\n\n22 NOV 06, 0420 hrs, GMT.
226471
get it sorted 1st, \nthen get the differences of adjacent number pairs, \nget the min of this differences
577652
Since you specified that the language is "C", it's\nworth pointing out that the Ansi standard guarantees\nthat the implementation will include a "qsort" function.\n\nAll you need to do is write the simple comparison\nfunction which is passed to qsort, and you're done.\n\nI copied the following (including the little test program)\ndirectly from the qsort man page:\n\n#include <stdlib.h>\n#include <stdio.h>\n\nstatic int\nintcompare(const void *p1, const void *p2)\n{\n  int i = *((int *)p1);\n  int j = *((int *)p2);\n\n  if (i > j)\n    return (1);\n  if (i < j)\n    return (-1);\n  return (0);\n}\n\nint\nmain()\n{\n  int i;\n  int a[10] = { 9, 8, 7, 6, 5, 4, 3, 2, 1, 0 };\n  size_t nelems = sizeof (a) / sizeof (int);\n\n  qsort((void *)a, nelems, sizeof (int), intcompare);\n\n  for (i = 0; i < nelems; i++)\n    (void) printf("%d ", a[i]);\n\n  (void) printf("&#92;n");\n  return (0);\n}\n\n\nSee the references for the man page as well as the\nsource from the OpenSolaris site.
448114
In C++ you should do this:
997805
==================\n#include <iostream.h>\n\nint main(void)\n{\n\n   int count; // holds the count\n   float sum, number, avg;\n\n   sum = 0.0;\n   count = 0;\n\n   cout << endl << endl   //  Make some space\n        << endl << "Enter a number.  Enter a zero (0) when you have"\n        << endl << "finished entering numbers and want to compute the"\n        << endl << "average of the inputted numbers." ;\n\n   do {\n         cout << endl << endl << "Data element #" << count + 1\n              << ": ";\n\n         cin >> number;\n\n         if (number != 0.)\n           sum = sum + number;\n           count = count + 1;\n\n      } while (number != 0.0);\n\n      averg = sum / (count-1);\n      cout << endl << "The average is: " << avg\n           << endl << endl ;\n      return 0;\n}\n\n\n==================\n\nThis program computes the average of a set of data values. Commented a bit for you to understand\n\nEnjoy
538436
Copy the code here: http://www.ozgrid.com/VBA/Sort.htm and follow the instructions in the middle of the page that begin: "To use this UDF push Alt+F11..."
839112
Hold down the ALT key and press #3 on the numpad ♥