51nod1109(bfs)
题目链接:http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1109
题意:中文题诶~
思路:可以用二叉树构建,根节点为 1,左儿子为 0,右儿子为 1.然后直接bfs一遍就好了;
注意:直接用十进制记录可能会tle或mle.可以用二进制形式记录,再存储到十进制数里,输出时再还原成二进制形式就好了;
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <queue> 3 #define ll long long 4 using namespace std; 5 6 int n; 7 ll ans; 8 string res; 9 10 void bfs(void){ 11 queue<pair<int,int> > q; 12 q.push({1, 1}); 13 ll cnt, cnt1, cnt2; 14 while(!q.empty()){ 15 cnt = q.front().first; 16 int x = q.front().second; 17 q.pop(); 18 cnt1 = cnt * 2; 19 cnt2 = cnt * 2 + 1; 20 if((x * 10) % n == 0){ 21 ans = cnt1; 22 break; 23 } 24 if((x * 10 + 1) % n == 0){ 25 ans = cnt2; 26 break; 27 } 28 q.push({cnt1, (x * 10) % n}); 29 q.push({cnt2, (x * 10 + 1) % n }); 30 } 31 while(ans){ 32 res += (ans % 2 + '0'); 33 ans /= 2; 34 } 35 for(int i = res.size()-1; i >= 0; i--){ 36 cout << res[i]; 37 } 38 cout << endl; 39 } 40 41 int main(void){ 42 cin >> n; 43 bfs(); 44 }
我就是我,颜色不一样的烟火 --- geloutingyu