Uva537-artificial intelligence?

Calculate the key characters I, P, u in the given string.

Train of Thought: Self-experience is a bit painful to input ..... Enter two parts. For details, seeCode!!!!!

# Include <stdio. h> # include <string. h> int main () {int N, T = 0; char s [100]; scanf ("% d", & N); While (n --) {int CNT = 0; Double X, Y; char STR [2], ss [2]; memset (S, 0, sizeof (s )); while (S [CNT] = getchar ())! = ') CNT ++; STR [0] = s [CNT-1]; scanf ("% lf", & X); CNT ++; s [CNT] = getchar (); SS [0] = s [CNT]; if (s [CNT] = 'M' | s [CNT] = 'K' | s [CNT] = 'M ') {If (s [CNT] = 'M') x/= 1000; If (s [CNT] = 'k') x * = 1000; if (s [CNT] = 'M') x * = 1000000; CNT ++;} while (S [CNT] = getchar ())! = ') CNT ++; STR [1] = s [CNT-1]; scanf ("% lf", & Y); CNT ++; s [CNT] = getchar (); SS [1] = s [CNT]; if (s [CNT] = 'M' | s [CNT] = 'K' | s [CNT] = 'M ') {If (s [CNT] = 'M') y/= 1000; If (s [CNT] = 'k') y * = 1000; if (s [CNT] = 'M') y * = 1000000; CNT ++;} printf ("problem # % d \ n", ++ t ); if (STR [0] = 'U' & STR [1] = 'I ') | (STR [0] = 'I' & STR [1] = 'U') printf ("P = %. 2fw \ n ", x * Y); If (STR [0] = 'P' & STR [1] = 'U') printf (" I = %. 2fa \ n ", X/(y * 1.0); If (STR [0] = 'U' & STR [1] = 'P ') printf ("I = %. 2fa \ n ", Y/(x * 1.0); If (STR [0] = 'P' & STR [1] = 'I ') printf ("u = %. 2fv \ n ", X/(y * 1.0); If (STR [0] = 'I' & STR [1] = 'P ') printf ("u = %. 2fv \ n ", Y/(x * 1.0); printf (" \ n ");} return 0 ;}