HDU Computer Transformation1041 題解

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Problem DescriptionA sequence consisting of one digit, the number 1 is initially written into a computer. At each successive time step, the computer simultaneously tranforms each digit 0 into the sequence 1 0 and each digit 1 into the sequence 0 1. So, after the first time step, the sequence 0 1 is obtained; after the second, the sequence 1 0 0 1, after the third, the sequence 0 1 1 0 1 0 0 1 and so on. 

How many pairs of consequitive zeroes will appear in the sequence after n steps? 
 
InputEvery input line contains one natural number n (0 < n ≤1000). 
OutputFor each input n print the number of consecutive zeroes pairs that will appear in the sequence after n steps.
 
Sample Input

23

 
Sample Output

11

找規律的問題，分開0和1的衍生問題就好理解了。

#include <stdio.h>#include <vector>#include <string.h>#include <algorithm>#include <iostream>#include <string>#include <limits.h>#include <stack>#include <queue>#include <set>#include <map>using namespace std;const short MAX_N = 1001;vector<short> tbl0[MAX_N], tbl1[MAX_N];//inverse saved numbersvoid addLargeNum(vector<short> &rs, vector<short> &a, vector<short> &b){int n = (int)a.size(), m = (int)b.size();rs.clear();short carry = 0;for (int i = 0, j = 0; i < n || j < m || carry; i++, j++){short an = i < n ? a[i] : 0;short bn = j < m ? b[i] : 0;carry += an + bn;rs.push_back(carry % 10);carry /= 10;}}void plusOne(vector<short> &a){if (a.empty()) a.push_back(1);else{int i = 0;while (i < (int)a.size() && 9 == a[i]) a[i++] = 0;if (i == (int)a.size()) a.push_back(1);else a[i]++;}}void getTbl(){tbl0[0].push_back(0); tbl0[1].push_back(0);tbl1[0].push_back(0); tbl1[1].push_back(0);for (int i = 2; i < MAX_N; i++){addLargeNum(tbl0[i], tbl1[i-1], tbl0[i-1]);addLargeNum(tbl1[i], tbl0[i-1], tbl1[i-1]);if (!(i&1)) plusOne(tbl1[i]);}}int main(){getTbl();int n;while (scanf("%d", &n) != EOF){vector<short> &a = tbl1[n];short m = (short)tbl1[n].size();for (short i = m-1; i >= 0; i--){printf("%d", a[i]);}putchar('\n');}return 0;}

HDU Computer Transformation1041 題解