POJ3982: Sequence

The Description Series A meets the requirements of An = An-1 + An-2 + An-3, n> = 3 programming, given A0, A1, and A2, computing A99Input input contains multiple rows of data each row contains three integers A0, A1, A2 (0 <= A0, A1, A2 <= 32767) data ends with EOF Output. For each input row, Output A99 value Sample Input1 1 1 Sample output69087442470169109923566147 [cpp] # include <stdio. h> # include <string. h> # include <stdlib. h> void add (char a [], char B [], char back []) {int I, j, k, up, x, y, z, l; char * c; if (strlen (a)> strlen (B) l = strlen (a) + 2; Else l = strlen (B) + 2; c = (char *) malloc (l * sizeof (char); I = strlen (a)-1; j = strlen (B)-1; k = 0; up = 0; while (j> = 0 | I> = 0) {if (I <0) x = '0'; else x = a [I]; if (j <0) y = '0'; else y = B [j]; z = x-'0' + y-'0'; if (up) z ++; if (z> 9) {up = 1; z % = 10 ;} else up = 0; c [k ++] = z + '0'; I --; j --;} if (up) c [k ++] = '1 '; I = 0; c [k] = '\ 0'; for (k-= 1; k> = 0; k --) back [I ++] = c [k]; back [I] = '\ 0';} int main (){ Char a [1000], B [1000], c [1000], sum [1000]; int m, I; while (~ Scanf ("% s", a, B, c) {add (a, B, sum); add (c, sum, sum ); int n = 5; for (I = 1; I <= 24; I ++) {add (B, c, a); add (sum, a, ); add (c, sum, B); add (B, a, B); add (sum, a, c); add (B, c, c); add (, b, sum); add (c, sum, sum);} printf ("% s \ n", sum);} return 0 ;}