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The 2018 ACM-ICPC Asia Qingdao Regional Contest

时间:2018-11-11 14:58:45      阅读:217      评论:0      收藏:0      [点我收藏+]

标签:icpc   ace   line   strlen   target   main   contest   typedef   efi   

The 2018 ACM-ICPC Asia Qingdao Regional Contest

青岛总体来说只会3题

C

#include<bits/stdc++.h>
using namespace std;
#define maxn 3000005
char a[maxn],b[maxn];
int c[maxn],ll[maxn],rr[maxn];
int main(){
   int t;
   cin>>t;
   while(t--){
      int n;
      scanf("%d",&n);
      scanf("%s",a);
      scanf("%s",b);
      for(int j=0;j<n;j++){
         if(a[j]!=b[j]){
            c[j]=1;
         }else c[j]=0;
      }
      int i=0;
      bool fa=0;
      for(int j=0;j<strlen(a);j++){
         //if(fa&&c[j]==1) continue;
         if(fa==1&&c[j]==0){
            rr[i-1]=j;
            fa=0;
         }else if(fa==0&&c[j]==1){
            ll[i]=j;
            i++;
            fa=1;
         }
      }
      if(c[n-1]==1) rr[i-1]=n-1;
      if(i==0){
          printf("%lld\n",1LL*n*(n+1)/2);
      }else if(i==1){
          int k=rr[i-1]-ll[i-1]+1;
          long long ans=1LL*(k-1)*2+1LL*(n-k)*2;
          printf("%lld\n",ans);
      }else if(i==2){
         printf("6\n");
      }else{
        printf("0\n");
      }
   }
   return 0;
}

E

#include<bits/stdc++.h>
using namespace std;
#define maxn 200005
#define LL long long
LL a[maxn],b[maxn];
LL n,m;
bool fa(LL x){
  LL ans=0;
  for(int j=0;j<n;j++){
     if(x%a[j]==0){
        b[j]=x/a[j];
     }else{
        b[j]=x/a[j]+1;
     }
  }
  for(int j=0;j<n;j++){
     if(j==n-1&&b[j]<1) break;
     ans++;
     if(b[j]>1){
        ans+=(b[j]-1)*2;
        b[j+1]-=b[j]-1;
     }
     if(ans>m) return 0;
  }
  return 1;
}
int main(){
   int t;
   cin>>t;
   while(t--){
      scanf("%lld%lld",&n,&m);
      LL mx=0;
      for(int j=0;j<n;j++){
         scanf("%lld",&a[j]);
         mx=max(mx,a[j]);
      }
      LL l=0,r=mx*m;
      if(m==0) cout<<"0"<<endl;
      else{

         while(l<r){
           LL mid=(r+l)/2;
           if(fa(mid)) l=mid+1;
           else r=mid;
         }  
         printf("%lld\n",max(1LL*0,l-1));
      }
    
   }
   return 0;
}

J

#include<bits/stdc++.h>
using namespace  std;
typedef long long LL;
const int maxn = 100010;
int Min( int a , int b ){ return a<b?a:b; }
int n,m,a[maxn];
int main()
{
    int T; scanf( "%d" , &T );
    for( int cas=1 ; cas<=T ; cas++ )
    {
        int cnt = 0,num = 0;
        scanf ( "%d%d" , &n , &m );
        for( int i=1 ; i<=n ; i++ )
        {
            int x; scanf( "%d" , &x );
            if ( x==0 )  cnt++;
            else  a[++num] = x;
        }
        if ( m==n ) printf( "Richman\n" );
        else if ( m<cnt ) printf( "Impossible\n" );
        else
        {
            LL ans = 0; int tmp = 2000000000;
            for( int i=1 ; i<=num ; i++ )
            {
                if ( i<=m-cnt )  ans += a[i];
                else tmp = Min( tmp , a[i] );
            }
            printf( "%lld\n" , ans+tmp-1 );
        }
    }
    return 0;
}

 

M

#include<bits/stdc++.h>
using namespace std;
#define LL long long
int fa[12]={1,0,0,0,1,0,1,0,2,1};
int main(){
  int t;
  cin>>t;
  while(t--){
     int  n,z;
     scanf("%d%d",&n,&z);
     //cout<<"2"<<endl;
     while(1){
        if(n==1||n==0) break;
        if(z<=0) break;
         int i=0;
         int k=n;
         while(k>0){
            LL  ii=k%10;
            i+=fa[ii];
            k/=10;
         }
         n=i;
         z--;

     }
     if(z==0){
        printf("%d\n",n);
     }else{
        if(z%2==0){
           printf("%d\n",n);
        }else{
           printf("%d\n",fa[n]);
        }
     }
  }
}

 

The 2018 ACM-ICPC Asia Qingdao Regional Contest

标签:icpc   ace   line   strlen   target   main   contest   typedef   efi   

原文地址:https://www.cnblogs.com/DyLoder/p/9941973.html

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