Codeforces Round #450 (Div. 2) A题 + B题

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不是题解只是简单的记录下自己写过的东西~~

第一题其实主要是看你阅读水平的

给你一堆点然后问你能不能删掉一个点然后使其余所有的点都在Y轴的一侧很简单对吧~~

#include <iostream>#include <cstring>using namespace std;int main(){int _ , Left , Right, x , y ;Left = Right = 0;cin >>_;while (_ --) {cin >> x >> y;if (x > 0) Right ++;else Left ++;}if (Left <= 1 || Right <= 1) cout << "Yes" << endl;else cout << "No" << endl;}

第二题就比较有意思了

大概意思是这样的给你三个数字 a b c 问你a除以b的小数部分第几位是c 如果没有就输出-1

当时我没有想到模拟小学生除法的这种思想但是水过去了

做法很多找寻环节如果没找到就GG 对吧但是找的过程中有一个鸽巢原理循环节不可能比本身还长所以复杂度O(b)

这是出题人给的题解：

In this task you should complete long division and stop, when one period passed. Period can't be more than b by pigeonhole principle. So you need to complete b iterations and if c digit hasn't been met, print - 1.

Time complexity O(b).

我比赛时候的过法就比较神奇了：

java 大小数三百位暴力找没有就GG

先帖一个我的非主流过法：

import java.io.*;import java.util.*;import java.math.BigDecimal;public class Main {public static void main(String[] args) throws Exception {Scanner stdin = new Scanner(System.in);double a , b;//a = a - int(a / b) * b;Integer k;a = stdin.nextDouble();b = stdin.nextDouble();k = stdin.nextInt();BigDecimal aa = new BigDecimal(Double.toString(a));BigDecimal bb = new BigDecimal(Double.toString(b));String str = aa.divide(bb , 300, BigDecimal.ROUND_HALF_UP).toString();boolean flag = true;//System.out.println(str);int Ca = 1;for (int i = 2; i < 300; i ++) {String temp = String.valueOf(k);String tt = String.valueOf(str.charAt(i));//System.out.println("temp == " + temp + "tt == " +tt);//System.out.println("temp == " + str.charAt(i));if (temp.equals(tt)) {System.out.println(Ca);flag = false;break;}Ca = Ca + 1;}if (flag == true) System.out.println(-1);}}

然后贴个我模拟小学生除法过的代码：

#include <iostream>#include <cstring>using namespace std;const int N = 100005;int main(){ios::sync_with_stdio(0);cin.tie(0) , cout.tie(0);int a , b , c, temp, Case = 0;cin >> a >> b >> c;temp = a;memset(idx , false , sizeof(N));while (Case < 100000) {Case++;temp *= 10;int cur = temp / b;temp %= b;if (cur == c) {cout << Case <<endl;return 0;}}cout << -1 << endl;}

最后也是最犀利的解法是找循环节 tie一个cf大佬的解法吧：

人家这个写的真的简洁~（毕竟红名ORZ）

#include <cstdio>#include <iostream>#include <vector>#include <set>#include <map>#include <cmath>#include <string>#include <cstring>#include <sstream>#include <queue>#include <iomanip>#include <algorithm>using namespace std;const int Maxn = 1000005;int a, b, c;bool was[Maxn];int main(){cin >> a >> b >> c;for (int i = 1; ; i++) {a *= 10;if (was[a]) break;was[a] = true;int cur = a / b; a %= b;if (cur == c) { printf("%d\n", i); return 0; }}printf("-1\n");return 0;}

PS：写cf以来第一次+88 开心的不行 ~~

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