原題傳送門

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題面

Traveling Salesman Problem

time limit per test2 secondsmemory limit per test64 megabytesinputstandard inputoutputstandard output

problem descride

Nowadays British scientists has been investigating the opportunities to increase the performance of traditional computers. Recent research has shown that changing logical zero to logical right shift ( x = x / 2) and logical one to logical negation ( x = 1 - x) will lead to a significant architecture enhancement. Thus, new architecture will be able to solve some NP hard problems. Assume that mentioned above basic operations will be new zero and new one, respectively.

For example, well known traveling salesman problem solved in the proposed architecture will be equivalent to obtaining number (fraction) $\frac{a}{2^b}$ from a number of one (initial state). Therefore, you are required to write a program (sequence of zeros and ones) to solve traveling salesman problem for given a and b.

Input

Single line contains two integer numbers a and b.
$1?\le ?b?\le ?60$,
$1?\le ?a?<?2^b$,
a is odd.

Output

Single line should contain a binary sequence, which is a solution for traveling salesman problem using new architecture. If there are multiple solutions, the minimal length program should be printed.

Sample Input

3 3

Sample Output

0010

題意描述

給定一個(gè)a和b，將初始值為1的數(shù)字，通過盡可能少的操作0（將數(shù)字變?yōu)樵瓉淼?span id="ozvdkddzhkzd" class="katex--inline">$\frac{1}{2}$）和操作1（將數(shù)字變?yōu)?-原來的數(shù)字），變成值為$\frac{a}{2^b}$

題目分析

一開始以為和目標(biāo)數(shù)字的二進(jìn)制有關(guān)，但題目其實(shí)保證了目標(biāo)數(shù)字絕對(duì)能通過所給的兩個(gè)操作算出來，所以逆向思維。從目標(biāo)數(shù)字出發(fā)，若當(dāng)前數(shù)字大于$\frac{1}{2}$則說明當(dāng)前數(shù)字是由某個(gè)數(shù)字通過操作1之后得到的，因?yàn)榻?jīng)過操作1和操作0的數(shù)字必定小于1。如果通過操作0而得到一個(gè)大于$\frac{1}{2}$的數(shù)字，那么原數(shù)字則大于1。這是顯然錯(cuò)誤的。若當(dāng)前數(shù)字小于$\frac{1}{2}$，則說明當(dāng)前數(shù)字是通過操作0得到的。從而逆推出正確的操作順序。

具體代碼

#include <iostream> #include <algorithm> #include <vector> #include <map>using namespace std;const int maxn = 1e6 + 7;long long a, b, cnt; int ans[maxn];long long qpow(long long a, long long x) {long long res = 1;while (x){if(x&1) res *= a;a = a * a;x /= 2;}return res; }int main() {cnt = 0;cin >> a >> b;b = qpow(2, b);while(a != b){if (2 * a > b){a = b - a;ans[++cnt] = 1;}else{b /= 2;ans[++cnt] = 0;}}for (int i = cnt; i >= 1; i--) cout << ans[i];return 0; }

總結(jié)

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