Zoj 1201 Inversion

Zoj 1201 Inversion

P 5 9 1 8 2 64 7 3

I 2 3 6 4 0 22 1 0

I think this is a problem on the Internet, but I am really painstaking ~ So I need to write down my own well-behaved methods, which is better than none ~

The conversion from P to I is very simple. Find the number "1" in P. There are two numbers on the left of P, so the first number of I is 2, and so on ~

I to P conversion, key! For example, if the number 3 in I indicates that there are three numbers in front of 2 that are larger than him, then we need to see if the number 1 is in front of him, if it is in front of him or in its position (less than or equal to), the position of this number will be pushed back to a unit, but it is not enough to check it again, for example, for the 5th digits in I, it must be checked four times so that it has a sufficient opportunity to push back each digit before him. For example, the number "1" in I indicates the position of 8 in P, that is, subscript 3. At this time, the previous positions of 2, 3, 6, 4, 0, and 2 have been changed
2, 4, 8, 6, 0, 5, and 7. Then, when 1 encounters 0, it turns to 2, and then throws 0 to the array of numbers that have been determined by the definition. The second time, 2. When 2 is changed to 3, check it again several times, but there are no numbers smaller than or equal to 3, and 3 is no longer added, SO 3 is the subscript of number 8.

Code:

# Include <stdio. h> int main () {int I, j, n, m, K, G, n, a [1000], B [1000], L, C [1000], x, F, R, S [1000], Z; char t; for (I = 1; I ++) {scanf ("% d", & N ); if (n = 0) break; getchar (); scanf ("% C", & T); For (j = 0; j <= N-1; j ++) {scanf ("% d", & A [J]);} If (t = 'P') {for (j = 0; j <= N-1; j ++) {for (k = 0; k <= N-1; k ++) {if (a [k] = J + 1) break;} m = 0; for (G = 0; G <= K; G ++) {if (a [g]> J + 1) m ++;} If (J! = 0) printf (""); printf ("% d", m);} printf ("\ n");} If (t = 'I ') {for (j = 0; j <= N-1; j ++) {x = 1; F = 0; while (x <= J + 1) {for (G = 0; G <= J-1; G ++) {r = 1; for (Z = 0; Z <= F-1; Z ++) {If (G = s [Z]) {r = 0; break ;}} if (a [g] <= A [J] & R = 1) {s [f] = g; A [J] ++; // push the corresponding numeric subscript F ++ ;}} x ++ ;} B [A [J] = J + 1;} For (j = 0; j <= N-2; j ++) {for (k = J + 1; k <= N-1; k ++) {if (a [J]> A [k]) {L = A [k]; A [k] = A [J]; A [J] = L ;}}for (j = 0; j <= N-1; j ++) {If (J! = 0) printf (""); printf ("% d", B [A [J]) ;}} return 0 ;}