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Problem - A - Codeforces

查有没有人说谎，有一个必错的情况：

两个人都说输了，必有人撒谎，还有就是所有人都赢了，也是撒谎

来看代码：

#include <iostream>
#include <vector>
using namespace std;int main() 
{int t;cin >> t;while (t--){int n;cin >> n;vector<int>arr(n);int sum = 0;for (int i = 0; i < n; i++){cin >> arr[i];sum += arr[i];}if (sum == n){cout << "YES" << "\n";continue;}int flag = 1;//大于3个 滑动窗口解决if (n > 2){for (int i = 0; i < n - 2; i++){int one = arr[i];int two = arr[i + 1];int thr = arr[i + 2];//枚举过程if (one == 0 && two == 0){flag = 0;}if (two == 0 && thr == 0){flag = 0;}}}else{if (arr[0] == arr[1]){flag = 0;}}if (flag){cout << "NO" << "\n";}else{cout << "YES" << "\n";}}
}

Problem - B - Codeforces

如何躲得最好呢？

第一次切割：横切竖切有2种选择，我之前比较看谁切得多就切，但是会出错，所以我横竖都切一次，最后来比较

每次都躲中间，这个切就像二分，来看代码：

#include <iostream>
#include <vector>
using namespace std;int main() 
{int t;cin >> t;int sum = 0;while (t--){sum++;int a, b, c, d;cin >> a >> b >> c >> d;//第一次判断 横切还是竖切int height = max(a - c, c - 1);int broad = max(b - d, d - 1);int left = 0;int right = a - 1-height;int count1 = 0;while (left < right) {int mid = (left + right) >> 1;right = mid;count1++;}right = b - 1;while (left < right) {int mid = (left + right) >> 1;right = mid;count1++;}left = 0;right = a - 1;int count = 0;while (left < right) {int mid = (left + right) >> 1;right = mid;count++;}right = b - 1 - broad;while (left < right) {int mid = (left + right) >> 1;right = mid;count++;}cout << min(count1,count)+1 << "\n";}
}

看看错误代码：

#include <iostream>
#include <vector>
using namespace std;int main() 
{int t;cin >> t;while (t--){int a, b, c, d;cin >> a >> b >> c >> d;//第一次判断 横切还是竖切int height = max(a - c, c - 1);int broad = max(b - d, d - 1);//如果是横切if (height * b > broad * a){a -= height;}else{b -= broad;}int left = 0;int right = a - 1;int count = 0;while (left < right) {int mid = (left + right) >> 1; right = mid; count++;}right = b - 1;while (left < right) {int mid = (left + right) >> 1;  right = mid;  count++;}cout << count+1 << "\n";}
}