PAT-1084（外观数列 ) && PAT-1085 (PAT单位排行)

1084  利用字符串string的可加性

 #include <bits/stdc++.h>
 using namespace std;
 int main ()
 {
     int x,n;
     cin>>x>>n;
     string str="x";
     str[]=''+x;
     for (int i=;i<=n;i++) {
         string  tmp="";
         string  t2="x";
         char k=str[]; int num=;
         for (int j=;j<str.size();j++) {
             if (str[j]==k) num++;
             else {
                 t2[]=k;       tmp+=t2;
                 t2[]=''+num; tmp+=t2;
                 k=str[j]; num=;
             }
         }
         t2[]=k;       tmp+=t2;
         t2[]=''+num; tmp+=t2;
         str=tmp;
     }
     cout<<str<<endl;
     return ;
 }

1085 利用map 实现学校名称的映射

 #include <bits/stdc++.h>
 using namespace std;
 const int N=1e5+;
 struct T {
     string name;
     double sum;
     int p;
     int num;
 };
 map < string,int > mp;
 T t[N];
 int cnt;
 int n;
 bool cmp (T x,T y) {
     if (x.p==y.p) {
         if (x.num==y.num)
             return x.name<y.name;
         return x.num<y.num;
     }
     return x.p>y.p;
 }
 int main ()
 {
     ios::sync_with_stdio(false);// 不加这句话会超时
     cin>>n;
     for (int i=;i<=n;i++) {
         string s1,s2; int val; double w;
         //getchar();// 很奇怪 ios::同步后就不需要加 而不同步就需要加
         cin>>s1>>val>>s2;
         for (int j=;j<s2.size();j++)
             if (s2[j]>='A'&&s2[j]<='Z')
                 s2[j]=s2[j]-'A'+'a';
         if (s1[]=='B')  w=(double)val/1.5;
         else if (s1[]=='A')  w=(double)val;
         else                  w=(double)val*1.5;
         if (!mp[s2]) {
             mp[s2]=++cnt;
             t[cnt].name=s2;
             t[cnt].num=;
             t[cnt].sum=w;
         }
         else {
             int id=mp[s2];
             t[id].num++;
             t[id].sum+=w;
         }
     }
     for (int i=;i<=cnt;i++)
         t[i].p=(int)t[i].sum;
     sort (t+,t++cnt,cmp);
     cout<<cnt<<"\n";

     int k=; int i=;  int point=t[].p;
     cout<<k<<" "<<t[i].name<<" "<<t[i].p<<" "<<t[i].num<<endl;
     for (i=;i<=cnt;i++) {
         if (t[i].p==point)
             cout<<k<<" "<<t[i].name<<" "<<t[i].p<<" "<<t[i].num<<endl;
         else {
             point=t[i].p;
             k=i;
             cout<<k<<" "<<t[i].name<<" "<<t[i].p<<" "<<t[i].num<<endl;
         }

     }
     return ;
 }