USACO sec1.1

感觉usaco的每道题都很不错，因为虽然简单但是都有点麻烦，很锻炼编程能力

1. Your Ride Is Here纯水题

   /*
   ID: wutao161
   PROG: ride
   LANG: C++     
   */
   #include <iostream>
   #include <fstream>
   #include <string>
   #include <string.h>
   using namespace std;
   
   char cmt[7],grp[7];
   int sc,sg;
   
   int main() {
       ofstream fout ("ride.out");
       ifstream fin ("ride.in");
       
       while(fin>>cmt>>grp)
       {
           sc=sg=1;
           for(int i=0;i<strlen(cmt);i++)
               sc=sc*(cmt[i]-64)%47;
           for(int i=0;i<strlen(grp);i++)
               sg=sg*(grp[i]-64)%47;
                 
           if( sc==sg )
               fout<<"GO"<<endl;
           else
               fout<<"STAY
   ";
       }
       return 0;
   }

2. Greedy Gift Givers水题

   #include<iostream>
   #include<fstream>
   #include<map>
   #include<string>
   #include<vector>
   #include<cstring>
   using namespace std;
   
   ifstream fin("gift1.in");
   ofstream fout("gift1.out");
   
   int n;
   string name[10];
   string t;
   int leftover[10];
   int ini[10];
   int final[10];
   
   
   int main(){
       
       map<string,int> s;
       int pos,m,p,gift;
       fin>>n;
       memset(ini,0,sizeof(ini));
       memset(final,0,sizeof(final));
       
       for(int i=0;i<n;i++)
       {
           fin>>name[i];
           s[name[i]]=i;
       }//½«ÈËÃûÓëÆäÐòºÅÒ»Ò»¶ÔÓ¦ 
       
       for(int k=0;k<n;k++){
           fin>>t;
           pos=s[t];
           fin>>m>>p;
           if(!p)
               continue;
           
           ini[pos]=m;
           final[pos] +=  m % p ;
           gift = m/p;
           for(int i=0;i<p;i++)
           {
               fin>>t;
               pos=s[t];
               final[pos]+=gift;
           }
       }
       
       for(int i=0;i<n;i++){
           fout<<name[i]<<" "<<final[i]-ini[i]<<endl;
       }
       
       return 0;
   }

3. Friday the Thirteenth（日历计算的水题）

   /*
   ID: wutao161
   PROG: friday
   LANG: C++         
   */
   
   #include<iostream>
   #include<fstream>
   #include<cstring>
   using namespace std;
   
   int norm[12]={31,28,31,30,31,30,31,31,30,31,30,31};
   int leap[12]={31,29,31,30,31,30,31,31,30,31,30,31};
   int day[7];//frequency  //monday is 1
   
   int main(){
       
       freopen("friday.in","r",stdin);
       freopen("friday.out","w",stdout); 
       int n;
       cin>>n;
   
       memset(day,0,sizeof(day));
       
       int s=0;//Ò»ºÅÊÇÐÇÆÚ¼¸      
       int thisyear;
       for(int i=0;i<n;i++)
       {
           thisyear=1900+i;
           if( thisyear % 100 == 0 )
           {
               if( thisyear%400==0 ) //leap
                   for(int j=0;j<12;j++){
                       day[(s+5)%7]++;
                       s+=(leap[j]-1)%7+1;
                       s%=7;
                   }
               else
                   for(int j=0;j<12;j++){
                       day[(s+5)%7]++;
                       s+=(norm[j]-1)%7+1;
                       s%=7;
                   }    
           }
           else if( thisyear % 4 != 0 )
           {
               for(int j=0;j<12;j++){
                   day[(s+5)%7]++;
                   s+=(norm[j]-1)%7+1;
                   s%=7;
               }
           }
           else //leap
           {
               for(int j=0;j<12;j++){
                   day[(s+5)%7]++;
               //    cout<<(s+5)%7<<" ";
                   s+=(leap[j]-1)%7+1;
                   s%=7;
               //    cout<<s<<"
   ";
               }
           }
       }
       
       for(int i=5;i<7;i++)    cout<<day[i]<<" ";
       for(int i=0;i<4;i++)    cout<<day[i]<<" ";
       cout<<day[4]<<endl;
       return 0;
   }

4. Broken Necklace(约瑟夫环变形  模拟)

   /*
   ID: wutao161
   PROG: beads
   LANG: C++11
   */
   #include<vector>
   #include<iostream>
   #include<cstring>
   #include<string>
   #include<algorithm>
   #include<cstdio>
   #include<set>
   #include<map>
   #include<cmath>
   using namespace std;
   
   int nex[350],pre[350];
   int n;
   bool vis[350];
   string s;
   
   void ini(){
       for(int i=0;i<n-1;i++)
           nex[i]=i+1;
       nex[n-1]=0;
       for(int i=1;i<n;i++)
           pre[i]=i-1;
       pre[0]=n-1;
   }
   
   int right(int st){
       
       int cnt=0,i=st;
       char bgcolor=s[st];
       while(s[i]==bgcolor || s[i]=='w' )
       {
           //vis[i]=1;
           if(cnt>=n)
               break;
           i=nex[i];
           cnt++;
           if(bgcolor=='w')
               bgcolor=s[i];
       }
       return cnt;
   }
   
   int left(int st){
       int cnt=0,i=st;
       char bgcolor=s[st];
       while(s[i]==s[st] || s[i]=='w')
       {
           //if(vis[i])
           //    break;
           if(cnt>=n)
               break;
           i=pre[i];
           cnt++;
           if(bgcolor=='w')
               bgcolor=s[i];
       }
       return cnt;
   }
   
   int main(){
       freopen("beads.in","r",stdin);
       freopen("beads.out","w",stdout);
       cin>>n;
       cin>>s;
       int ans=-1000;
       ini();
       for(int i=0;i<n-1;i++)
       {
           ans=max(ans,left(i)+right(i+1));
       }
       if(ans>n)
           ans=n;
       cout<<ans<<endl;
       return 0;
   }
   
   /*
   DESCRIPTION:
   可以利用约瑟夫环的链表思想来处理环 
   也可以用队列 
   */