Codeforces Round 912 (Div. 2)

A

大等于2依据冒泡排序即可排序，因此判断下1即可

#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10;
int a[N];void solve()
{int n , m;cin >> n >> m;for(int i = 1 ; i <= n ; i ++){cin >> a[i];}if(m < 2){for(int i = 2 ; i <= n ; i ++)if(a[i] < a[i - 1]){cout << "NO\n";return ;}}cout << "YES\n";return ;
}int main()
{int T;cin >> T;while(T --){solve();}return 0;
}

B

对每一个数字找哪一位肯定为0填0其他的填1

#include <bits/stdc++.h>using namespace std;const int N = 1e3 + 10;
int a[N] , m[N][N];void solve()
{int n;cin >> n;for(int i=1;i<=n;i++){a[i]=(1<<30)-1;for(int j=1;j<=n;j++){cin>>m[i][j];if(i!=j)a[i]&=m[i][j];}}for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){if((i!=j)&&((a[i]|a[j])!=m[i][j])){cout << "NO\n";return ;}}}cout << "YES\n";for(int i=1;i<=n;i++){cout << a[i] << " ";}    cout << endl;return ;
}int main()
{int T;cin >> T;while(T --){solve();}return 0;
}

C

从后往前考虑加到当前集合或新立一个集合

#include <bits/stdc++.h>using namespace std;const int N = 1e5 + 10;
long long a[N] , f[N];
int n;void solve()
{cin >> n;for(int i = 1 ; i <= n ; i ++){cin >> a[i];f[i]= -(1ll<<60);}f[n]=a[n];long long s=a[n];//加/不加for(int i=n-1;i;i--)f[i]=max(f[i+1]+a[i],f[i+1]+s+a[i]),s+=a[i];cout<<f[1]<<'\n';return ;
}int main()
{int T;cin >> T;while(T --){solve();}return 0;
}

D

从最高位开始考虑能否为1，计算能否时每个数当前位为1

#include<bits/stdc++.h>
#define ll long long
#define int long long
using namespace std;
const int N=2e6+1;
const int mo=998244353;
int a[N];
int x,y,n,sum,m,k,z,p,b[N],c[N];
void sovel()
{
cin>>n>>m;
for(int i=1;i<=n;i++)
{cin>>a[i];b[i]=a[i];
}
while(m--)
{cin>>y;int manx=0;for(int i=62;i>=0;i--){int x=(1LL<<i),sum=0;for(int j=1;j<=n;j++){if(((a[j]>>i)&1)==0){sum+=x-(a[j]%x);if(sum>y){break;}}}if(y>=sum){manx|=x;y-=sum;for(int j=1;j<=n;j++){if(((a[j]>>i)&1)==0){a[j]+=x-(a[j]%x);}}}}for(int i=1;i<=n;i++){a[i]=b[i];}cout<<manx<<endl;
}
}
signed main()
{
int t=1;
while(t--)
{sovel();
}
return 0;
} 

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