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[0]='0'+x;
    for (int i=2;i<=n;i++) {
        string  tmp="";
        string  t2="x";
        char k=str[0]; int num=1;
        for (int j=1;j<str.size();j++) {
            if (str[j]==k) num++;
            else {
                t2[0]=k;       tmp+=t2; 
                t2[0]='0'+num; tmp+=t2;
                k=str[j]; num=1;
            }
        }
        t2[0]=k;       tmp+=t2; 
        t2[0]='0'+num; tmp+=t2;
        str=tmp;
    }
    cout<<str<<endl;
    return 0;
}

1085 利用map 实现学校名称的映射

#include <bits/stdc++.h>
using namespace std;
const int N=1e5+3;
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=1;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=0;j<s2.size();j++) 
            if (s2[j]>='A'&&s2[j]<='Z') 
                s2[j]=s2[j]-'A'+'a';
        if (s1[0]=='B')  w=(double)val/1.5;
        else if (s1[0]=='A')  w=(double)val;
        else                  w=(double)val*1.5;
        if (!mp[s2]) {
            mp[s2]=++cnt;
            t[cnt].name=s2;
            t[cnt].num=1;
            t[cnt].sum=w;
        }
        else {
            int id=mp[s2];
            t[id].num++;
            t[id].sum+=w;
        }
    }
    for (int i=1;i<=cnt;i++)
        t[i].p=(int)t[i].sum;
    sort (t+1,t+1+cnt,cmp);
    cout<<cnt<<"
";

    int k=1; int i=1;  int point=t[1].p;
    cout<<k<<" "<<t[i].name<<" "<<t[i].p<<" "<<t[i].num<<endl;
    for (i=2;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 0;
}