• Codeforces Round #619 (Div. 2)


    A.Three Strings

     

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 typedef long long ll;
     4 const int N = 1e5 + 7;
     5 const int inf = 0x3f3f3f3f;
     6 char a[110],b[110],c[110];
     7 int t;
     8 int main()
     9 {
    10     scanf("%d",&t);
    11     while(t--)
    12     {
    13         scanf("%s%s%s",a,b,c);
    14         int len=strlen(a);
    15         bool ok=false;
    16         for(int i=0;i<len;i++)
    17         {
    18             if(a[i]==c[i]||b[i]==c[i]) continue;
    19             ok=true;
    20         }
    21         if(ok) puts("NO");
    22         else puts("YES");
    23     }
    24     return 0;
    25 }
    View Code

    B. Motarack's Birthday

     

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 typedef long long ll;
     4 const int N = 1e5 + 7;
     5 const int inf = 0x3f3f3f3f;
     6 int a[N];
     7 int t,n;
     8 void work(int a[])
     9 {
    10     bool ok=true;
    11     int maxx=0,minx=inf;
    12     int res=0;
    13     for(int i=1;i<=n;i++)
    14     {
    15         if(a[i]!=-1)
    16         {
    17             ok=false;
    18             if(a[i-1]!=-1) res=max(res,abs(a[i]-a[i-1]));
    19             if(a[i+1]!=-1) res=max(res,abs(a[i]-a[i+1]));
    20             continue;
    21         }
    22         if(a[i-1]!=-1) maxx=max(maxx,a[i-1]),minx=min(minx,a[i-1]);
    23         if(a[i+1]!=-1) maxx=max(maxx,a[i+1]),minx=min(minx,a[i+1]);
    24     }
    25     if(ok)
    26     {
    27         cout<<0<<" "<<1<<endl;
    28  
    29     }
    30     else
    31     {
    32         int k = (maxx+minx+1)/2;
    33         cout<<max(res,max(abs(maxx-k),abs(minx-k)))<<" "<<k<<'
    ';
    34     }
    35 }
    36 int main()
    37 {
    38     scanf("%d",&t);
    39     while(t--)
    40     {
    41         scanf("%d",&n);
    42         for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    43         a[0]=-1, a[n+1]=-1;
    44         work(a);
    45     }
    46     return 0;
    47 }
    View Code

    C. Ayoub's function

     

     1 #include <bits/stdc++.h>
     2 using namespace std;
     3 typedef long long ll;
     4 const int N = 1e5 + 7;
     5 const int inf = 0x3f3f3f3f;
     6 int T;
     7 ll n,m;
     8 void work(ll n, ll m)
     9 {
    10     ll ans=(n+1)*n/2;
    11     ll d=n-m;
    12     ll x=d/(m+1);
    13     ll y=d%(m+1);
    14     ll cnt=m-y+1;
    15     ans-=cnt*(x*(x+1)/2);
    16     ans-=y*((x+2)*(x+1)/2);
    17     cout<<ans<<'
    ';
    18 }
    19 int main()
    20 {
    21     scanf("%d",&T);
    22     while(T--)
    23     {
    24         cin>>n>>m;
    25         work(n,m);
    26     }
    27     return 0;
    28 }
    View Code
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  • 原文地址:https://www.cnblogs.com/Edviv/p/12308037.html
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