Problem - A - Codeforces

AC代码:

#include<bits/stdc++.h>
#define endl '\n'
//#define int long long
using namespace std;
const int N=1e3+10;
int a[N];
int n;
void solve() {cin>>n;int ans=0;for(int i=1;i<=n;i++) cin>>a[i],ans^=a[i];if(n%2==1){for(int i=0;i<(1<<8);i++){if((ans^i)==0){cout<<i<<endl;return;}}cout<<-1<<endl;}else{if(ans==0) cout<<0<<endl;else cout<<-1<<endl;}
}
int main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int t=1;cin>>t;while(t--) {solve();}return 0;
}

Problem - B - Codeforces 

长度为n的字符串
操作有且只有一次，刚好一次
操作：选择一个字符，将它放在开头

输出字典序最小的字符串

先找到最小的字符是哪一个，然后从后往前遍历，找到第一个和最小字符相同的，将它放在开头

AC代码:

#include<bits/stdc++.h>
#define endl '\n'
//#define int long long
using namespace std;
int n;
string s;
void solve() {cin>>n;cin>>s;char minn='z';for(int i=0;i<(int)s.size();i++){if(minn>s[i]) minn=s[i];}int idx=-1;for(int i=(int)s.size()-1;i>=0;i--){if(s[i]==minn){idx=i;break;}}cout<<s[idx];for(int i=0;i<(int)s.size();i++){if(i==idx) continue;cout<<s[i];}cout<<endl;
}
int main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int t=1;cin>>t;while(t--) {solve();}return 0;
}

Problem - C - Codeforces

 数形结合，从图中可以看出，找最接近切线的两条线，判断是否和抛物线有交点即可（用求根公式）

注意：当取pos-1时，pos的范围应在1到s.size()

 AC代码:

#include <bits/stdc++.h>
#define endl '\n'
#define int long long
using namespace std;
int n, m;
int k;
void solve() {cin >> n >> m;vector<int>ans;while (n--) {cin >> k;ans.push_back(k);}sort(ans.begin(), ans.end());while (m--) {int a, b, c;cin >> a >> b >> c;double k1 = b + 2 * sqrt(a * c);double k2 = b - 2 * sqrt(a * c);int pos = lower_bound(ans.begin(), ans.end(), k1) - ans.begin();int pos1 = lower_bound(ans.begin(), ans.end(), k2) - ans.begin();if (pos>=0&&pos<(int)ans.size()&&(b - ans[pos]) * (b - ans[pos])-4*a*c< 0) {cout << "Yes" << endl;cout << ans[pos] << endl;} else if (pos>0&&pos<(int)ans.size()+1&& (b - ans[pos - 1]) * (b - ans[pos - 1]) - 4 * a * c < 0) {cout << "Yes" << endl;cout << ans[pos - 1] << endl;}else if (pos1>=0&&pos1<(int)ans.size()&&(b - ans[pos1]) * (b - ans[pos1])-4*a*c < 0) {cout << "Yes" << endl;cout << ans[pos1] << endl;} else if (pos1>0&&pos1<(int)ans.size()+1&& (b - ans[pos1 - 1]) * (b - ans[pos1 - 1]) - 4 * a * c < 0) {cout << "Yes" << endl;cout << ans[pos1 - 1] << endl;}else{cout<<"No"<<endl;}}cout << endl;
}
signed main() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);int t = 1;cin>>t;while (t--) {solve();}return 0;
}