please output this program printf("&#92;x30&#92;48");?
479514
It printed this for me 0♦8\n\nHope this Helps.\nGood Luck!!!
253461
®\nuse ascii:\n\n\n<p>& # 1 7 4 ;</p>\nwithout the spaces
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}
785470
There are numerous ways to do that. \nHere is, probably the simplest:\n\nchar *s="char *s=%c%s%c;%cmain() {printf(s,34,s,34,10,10);}%c";\nmain() {printf(s,34,s,34,10,10);}\n\nThe program itself is fairly useless.\nThe whole point of an exercise like this is the satisfaction of finding a solution yourself. Too bad, you did not give yourself a chance
482701
"mode"\n"display digits"\n"float 12" (or however many you want - 12 is the highest)
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}
169151
int\nmain (void)\n{\nfor(i=0;i<10;i++){\nprintf("some thing output here %d", i)\n}\n}
225777
From the DOS prompt, type Format c:
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}
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!
81883
Details of the algorithm for conversion along with\nsource program written in C can be found at.\nhttp://www.codepedia.com/1/Art_Expressions_p1\n\nA good short article on this topic.\nhttp://www.cs.man.ac.uk/~pjj/cs2121/fix.html
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.
608853
Code written in C and it works in C++\n\n#include <stdio.h>\n#include <conio.h>\n\nvoid main() {\n     float f[5];\n     int   i[5];\n     int j;\n     clrscr();\n     printf("Please, enter five float numbers: &#92;n");\n     for (j=0;j<5;j++){\n scanf("%f",&f[j]);\n i[j]=(int) f[j];\n     }\n     printf("Float value&#92;n");\n     for (j=0;j<4;j++)\n printf("%2.2f&#92;n",f[j]);\n     printf("&#92;nInteger value&#92;n");\n     for (j=0;j<4;j++)\n printf("%d&#92;n",i[j]);\n\n     getch();\n}
12600
In Custom format you can specify format as required by you.\nIt has three parts seperated by ;\n1st is for positive\n2nd is for negetive\n3rd is for Zero\nso you can specify as  follows to get the + sign displyed\n\n+#;-#;"Zero"
808564
printf("$%i.%i",dollars,cents);\n\nor if you have a floating point value you can do\n\nprintf("%5.2f",money);
423398
here...I'll tell you what to write and you write it...\n\n/* outline the program */\n\n// void main()...etc..\n\n/* define your variables */\n/* you'll need a pair of char pointers */\n/* and some numbers to hold converted values */\n\n\n/* input the first binary as a string */\n\n/* input the second binary as a string */\n\n/* convert them to decimal */\n\n/* add them */\n\n/* fprintf them */\n\n/* end program */\n\n\nthat's all.\nI write all my programs in commented logic first then fill in the blanks so that it does what I meant.  When I'm done, the program follows the flow and is already commented for the next person to follow.
400127
C  H   O\n.6  12   6\n\nWITHOUT DECIMAL POINT