Poj 1090 chain

A 0 1 string requires you to change all to 0. The change rule is that when the nth string is changed, the n-1 string must be 1, and all the values before n-1 must be 0, n> = 3; 1st can be changed at will, and 2nd must be changed when 1st is set to 1. How many times can we change the value of all the given 0 1 strings to 0?

First of all, I personally think about this type of question: when I look at the data and results so big that the search will definitely not work, it should be a regular formula question. When I think of hanruata, I should first look for a formula; Z [N] Function Representation

For 0 0 0 ...... The Nth Number of 0 is 1, and X [N] indicates 0 0 ...... The Nth number (1) of 1 0 is 0

Then there is Z [N] = Z [n-1] + 1 + x [n-1];… 1

     x[n]=z[n-1]+1+x[n-1];  ……2

Z [N] = 2 ^ n-1; X [N] = 2 ^ n-1;

A For loop can find the answer ans. The answer is stored in binary. The problem is mainly difficult to convert the binary into decimal because I feel that my thinking is correct, but the teacher times out, I am so angry that I have been making code changes, and I am confused by my own ideas. Therefore, the code quality is very poor. For details, see comments. Haha # include <stdio. h> # include <string. h> int array [1001]; int res [1001], ANS [1001]; // The binary string of ANS is saved. The decimal value converted from ANS is int INI () in res () {int Z, flag = 0; For (Z = 1; Z <= 1000; Z ++) {res [Z] + = flag; flag = res [Z]/10; Res [Z] % = 10;} return 0;} int add (int * flag, int fnum) {int I; for (I = 1; I <= fnum; I ++) RES [I] + = Flag [I]; return INI ();} int getres (Int n) {int I, j, Z, flag [1001], FF, fnum = 1; memset (flag, 0, sizeof (FLAG )); flag [1] = 1; for (I = 1; I <= N; I ++) {If (ANS [I]) add (flag, fnum ); for (j = 1; j <= fnum; j ++) {flag [J] * = 2;} FF = 0; For (Z = 1; Z <= fnum; Z ++) {flag [Z] + = ff; FF = Flag [Z]/10; flag [Z] % = 10;} while (FF! = 0) {flag [++ fnum] = FF % 10; FF/= 10 ;}return 0 ;}int main () {int I; int N, flag [1001], xx; while (scanf ("% d", & N )! = EOF) {memset (flag, 0, sizeof (FLAG); // the target State is 0 at the beginning; memset (Res, 0, sizeof (RES )); // initialize memset (ANS, 0, sizeof (ANS) for the decimal answer; // initialize the binary answer for (I = 1; I <= N; I ++) scanf ("% d", & array [I]); for (I = N; I> = 1; I --) // greedy forward from the back, for example, to solve the fourth different problem, you must change the first three States to 0 0 1 {If (flag [I] = array [I]). // if the problem is the same, no conversion is required; continue; flag [I-1] = 1; // if not the same, then you need to convert the previous 0 0 0 ...... 1 and then add step I to the same, and then add the first N-1 (0 0 0 ...... 1) The result is 0, and the formula is 2 ^ n-1, which corresponds to the n + 1 digit in the binary. ans [I] = 1 ;} // The binary calculation is followed by the conversion problem. You need to pay attention to it in one place, and there is nothing to mention about getres (n); xx = 0; for (I = 1000; i> = 1; I --) {If (xx = 0 & res [I] = 0) continue; xx = 1; printf ("% d ", res [I]);} If (xx = 0) printf ("0"); printf ("\ n");} return 0 ;}