how can i find first three output values for.....................?
966844
f(0) = -i\nf(1) = 1² - i = 1-i\nf(2) = 2² - i = 4-i\n\nI take it you are dealing with complex numbers?  i is the square root of -1.\n\nNRT
617746
sample space {1,2,3,4,5,6}
509467
1: x<12\n2: x>257\n3: x>42\n4: x>84\n5: x>25\n6: x>179\n7: x<8\n8: x>577\n9: x<113
1149797
The first one is (-5, 3). I don't remember how to do the others.
730496
iteration1: a = 3, b = 2\niteration2: a = 13, b = 8\niteration3: a = 51, b = 30\nloop exit:  a = 81, b = 30 // condition a = a + b executed before exit
605278
72201,02,03,04,05,06,07,09,10,11,12,14,15,16,17,19,21,22,23,25,27,31,60,95\n\n72213,14,15,16,17,18,19,20,24,90,98,99
1151802
fttft\n!. F 2. T 3. T 4. F 5. T
465985
1 - D\n2 - B\n3 - C\n4 - C
650616
These three sequntal clock wise numbers \n2+3+4 --> 9+4 --> 1 (adding the last no.(4)\nBut not works for all
894472
Use a calculator.  Enter [2] [log] for the first one.
830317
f1>f3>f2>f5>f4\nJust replace the variable n in  Excel's function and compare the result.
404639
you count the number of times each figure appears (this has been asked many other times):\nfirst row: 1\nsecond row: one 1: 11\nthird row: two 1: 21\nfourth row: one 2 one 1: 1211\nfifth row: one 1 one 2 two 1: 111221\nsixth row: three 1 two 2 one 1: 312211\n\n... and so on ...
1178151
If you know max.allowed width, use \nprintf( "%*.*s", width, width, arr[idx] )\n\nvoid printArray( char ** strings, int arr_sz, int columns, int min_width )\n{\nint rows = arr_sz/columns;\nfor( row = 0; row < rows, row++ )\nfor( col = 0; col < columns; col++ )\n{\n     int idx = columns*row+col;\n     if( idx >= arr_sz )\n           break;\n     printf( "%*.*s&#92;t", min_width, min_width , arr[idx] );\n}\nprintf( "&#92;n" );\n}
492690
S(I)+3f (g)--> SF (g)\n..........2...........6
966909
I am assuming that all you want is that simple pattern drawn.\n\nFor a predefined size of the triangle LINELIMIT:\n\nint main() {\n  int head = 1. prev; line = 0, col = 0;\n\n  for (line = 0; line < LINELIMIT; line++) {\n    prev = head;\n    for (col = 0; col < line; col++) {\n      printf("%d ", prev);\n      prev *= 2;\n    }\n\n    printf("&#92;n");\n    head *= 2;\n  }\n\n  return 0;\n};
633125
P, Q, R, S, T, U, V, W, X, Y, Z!!! Did I get it right???
865038
Do you mean printing the result of printf() statements?\n\nIf so, simply redirect the output of your program to either the printer directly or to a file and then print that file.\n\nEither on a linux box or from a command prompt on windows:\n\nc:&#92; myProgram > lpt1  or \nc:&#92; myProgram > myResults.txt