A - Creating Words
题意：略
代码：
#include<bits/stdc++.h>
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)//不能使用scanf了
#define int long long
#define loop(n) for(int i=0;i<n;i++)
#define rloop(n) for(int i=n-1;i>=0;i--)
#define print(c) cout<<"this "#c" is "<<c<<endl;
#define MAX INT_MAX
#define MIN INT_MIN
const int N=1010;using namespace std;
void solve() {string a,b;cin>>a>>b;char c=a[0];a[0]=b[0];b[0]=c;cout<<a<<' '<<b<<endl;
}
signed main() {ios;int n=1;cin >> n;while (n--)solve();
}
B - Maximum Multiple Sum
题意：略
代码：
#include<bits/stdc++.h>
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define int long long
#define loop(n) for(int i=0;i<n;i++)
#define rloop(n) for(int i=n-1;i>=0;i--)
#define print(c) cout<<"this "#c" is "<<c<<endl;
#define MAX INT_MAX
#define MIN INT_MIN
#define N 1e6+10
using namespace std;
void solve() {int n;cin>>n;auto check=[&](int t){int i=2;for(;i<=sqrt(t);i++)if(t%i==0)break;if(i>=sqrt(t))return 1;else return 0;};if(n<4)cout<<n<<endl;else cout<<2<<endl;
}
signed main() {ios;int n=1;cin >> n;while (n--)solve();
}
C - Good Prefixes
题意：给数组a[n]，然后定义数组的2*max=sum，则认为这个数组是好数组
我们求sum和max，然后比较即可，可以用递推优化
代码：
#include<bits/stdc++.h>
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define int long long
#define loop(n) for(int i=0;i<n;i++)
#define rloop(n) for(int i=n-1;i>=0;i--)
#define print(c) cout<<"this "#c" is "<<c<<endl;
#define MAX INT_MAX
#define MIN INT_MIN
#define N 1e6+10
using namespace std;
void solve() {int n; cin>>n;int a[n]; loop(n)cin>>a[i];int S[n],Max[n];loop(n){if(i==0)S[i]=a[i],Max[i]=a[i];else S[i]=S[i-1]+a[i],Max[i]=max(Max[i-1],a[i]);}int count=0;loop(n)if(S[i]==2*Max[i])count++;cout<<count<<endl;
}
signed main() {ios;int n=1;cin >> n;while (n--)solve();
}D - Manhattan Circle
题意：找圆的圆心
代码：
#include<bits/stdc++.h>
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define int long long
#define loop(n) for(int i=0;i<n;i++)
#define rloop(n) for(int i=n-1;i>=0;i--)
#define print(c) cout<<"this "#c" is "<<c<<endl;
#define MAX INT_MAX
#define MIN INT_MIN
#define N 1e6+10
using namespace std;
void solve() {int n,m;cin>>n>>m;char a[n][m];int line=n-1,num=0,first=0;for(int i=0;i<n;i++)for(int j=0;j<m;j++)cin>>a[i][j];for(int i=0,count=0,k=0;i<n;i++,count=0){for(int j=0;j<m;j++){if(a[i][j]=='#')count++;}if(count==2*k+1)k++;else if(k!=0){line=i-1,num=count==0?1:count+2;break;}//如果在最后一行，line不会赋值，则line的初始值设为n-1}loop(m)if(a[line][i]=='#'){first=i;break;}cout<<line+1<<' '<<first+num/2+1<<endl;
}
signed main() {ios;int n=1;cin >> n;while (n--)solve();
}