PAT Advanced Level (1081~1087)

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原创最新推荐文章于 2026-06-19 16:00:46 发布 1.4k 阅读 AI 写同款· GEO 优化 ›

PAT (Advanced Level) Practice python解法合集 1001 A+B Format 1002 A+B for Polynomials 1003 Emergency 1004 Counting Leaves 1005 Spell It Right 1006 Sign In and Sign Out 1007 Maximum Subsequence Sum 1008 Elevator 1009 Product of Polynomials 1010 R... 阅读详情

PAT 1081 Rational Sum

链接：http://www.patest.cn/contests/pat-a-practise/1081

代码：注意有总和为零的情况，此时要输出“0”

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<set>
#include<cmath>
#include<algorithm>
using namespace std;
long long gcd(long long x,long long b)
{
    long long c;
    for(;b!=0;)
    {
        c=x%b;
        x=b;
        b=c;
    }
    return x;
}
int main()
{
    int n;
    long long a,b,c,d,e;
    scanf("%lld",&n);
    scanf("%lld/%lld",&a,&b);
    for(int i=1;i<n;++i)
    {
        scanf("%lld/%lld",&c,&d);
        e=gcd(b,d);
        e=b/e*d;
        a=a*(e/b);
        c=c*(e/d);
        a=a+c;
        b=e;
    }
    long long f=a/b;
    a=a%b;
    if(f!=0||a==0)
    {
        printf("%lld",f);
    }
    if(a!=0)
    {
        c=gcd(a,b);
        if(f==0)
            printf("%lld/%lld",a/c,b/c);
        else
            printf(" %lld/%lld",a/c,b/c);
    }
    printf("\n");
    return 0;
}

PAT 1082 Read Number in Chinese

链接：http://www.patest.cn/contests/pat-a-practise/1082

代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<set>
#include<cmath>
#include<algorithm>
using namespace std;
string d[11]= {"ling", "yi", "er", "san", "si", "wu", "liu", "qi", "ba", "jiu"};
string q[11]= {"", "Shi", "Bai", "Qian", "Wan", "Shi", "Bai", "Qian", "Yi"};
vector<int> res;
vector<string> ans;
int main()
{
    int n;
    scanf("%d",&n);
    if(n==0)
    {
        printf("ling\n");
    }
    else
    {
        int len=0;
        if(n<0)
        {
            printf("Fu ");
            n=-n;
        }
        for(; n>0; n/=10)
        {
            res.push_back(n%10);
        }
        for(; len<res.size()&&res[len]==0; ++len);
        if(len<8)
        {
            for(int i=len; i<res.size(); ++i)
            {
                if(i!=0&&(res[i]!=0||i%4==0))
                {
                    ans.push_back(q[i]);
                }
                ans.push_back(d[res[i]]);
            }
            for(int i=ans.size()-1,j; i>=0; --i)
            {
                j=0;
                if(i!=ans.size()-1)
                    printf(" ");
                for(; i>=0&&ans[i]=="ling"; --i,++j);
                if(j>0&&ans[i]!="Wan")
                    printf("ling ");
                printf("%s",ans[i].c_str());
            }
            printf("\n");
        }
        else
        {
            printf("%s Yi\n",d[res[len]].c_str());
        }
    }
    return 0;
}

PAT 1083 List Grades

链接：http://www.patest.cn/contests/pat-a-practise/1083

代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<set>
#include<cmath>
#include<algorithm>
using namespace std;
struct student
{
    char name[15];
    char id[15];
    int grade;
} stu[100005];
bool cmp(const struct student &a,const struct student &b)
{
    return a.grade<b.grade;
}
int main()
{
    int n,a,b;
    int maxGrade=-1,minGrade=999;
    scanf("%d",&n);
    for(int i=0; i<n; ++i)
    {
        scanf("%s %s %d",stu[i].name,stu[i].id,&stu[i].grade);
        if(stu[i].grade>maxGrade) maxGrade=stu[i].grade;
        if(stu[i].grade<minGrade) minGrade=stu[i].grade;
    }
    sort(stu,stu+n,cmp);
    scanf("%d%d",&a,&b);
    if(a>b)
    {
        a=a^b;
        b=a^b;
        a=a^b;
    }
    if(b<minGrade||a>maxGrade)
    {
        printf("NONE\n");
    }
    else
    {
        int idxA=0;
        int idxB=n-1;
        for(;idxA<n&&stu[idxA].grade<a;++idxA);
        for(;idxB>=0&&stu[idxB].grade>b;--idxB);
        //printf("%d\n",idxB-idxA+1);
        for(int i=idxB;i>=idxA;--i)
            printf("%s %s\n",stu[i].name,stu[i].id);
    }
    return 0;
}

PAT 1086 Tree Traversals Again

链接：http://www.patest.cn/contests/pat-a-practise/1086

代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<set>
#include<stack>
#include<cmath>
#include<algorithm>
using namespace std;
struct point
{
    int sons[2];
} node[35];
bool first=true;
int n,m,root;
void postorder(int idx)
{
    for(int i=0; i<2; ++i)
        if(node[idx].sons[i]!=-1)
            postorder(node[idx].sons[i]);
    if(first)
    {
        printf("%d",idx);
        first=false;
    }
    else
        printf(" %d",idx);
}
int inorder()
{
    int id=-1;
    char s[10];
    if(m<2*n)
    {
        scanf("%s",s);
        if(strcmp(s,"Push")==0)
        {
            scanf("%d",&id);
            if(m==0) root=id;
            ++m;
        }
        else
        {
            ++m;
            return -1;
        }
        node[id].sons[0]=inorder();
        node[id].sons[1]=inorder();
    }
    return id;
}
int main()
{
    m=0;
    scanf("%d",&n);
    for(int i=1; i<=n; ++i)
    {
        node[i].sons[0]=node[i].sons[1]=-1;
    }
    inorder();
    postorder(root);
    return 0;
}

PAT 1087 All Roads Lead to Rome

链接：http://www.patest.cn/contests/pat-a-practise/1087

代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<set>
#include<map>
#include<cmath>
#include<algorithm>
using namespace std;
struct road
{
    vector<int> path;
    int costSum;
    int happinessSum;
}tmp;
map<int,string> idToName;
map<string,int> nameToId;
map<int,int> idToHappiness;
vector<int> path;
vector<struct road> ans;
bool visited[205];
int mapx[205][205];
int n;
void dfs(int st,int ed,int costSum,int happiniessSum)
{
    if(st==ed)
    {
        tmp.path.clear();
        int len=path.size();
        for(int i=0;i<len;++i)
            tmp.path.push_back(path[i]);
        tmp.costSum=costSum;
        tmp.happinessSum=happiniessSum;
        ans.push_back(tmp);
        return;
    }
    for(int i=0;i<n;++i)
    {
        if(mapx[st][i]>0&&(!visited[i]))
        {
            path.push_back(i);
            costSum+=mapx[st][i];
            happiniessSum+=idToHappiness[i];
            visited[i]=true;
            dfs(i,ed,costSum,happiniessSum);
            visited[i]=false;
            costSum-=mapx[st][i];
            happiniessSum-=idToHappiness[i];
            path.pop_back();
        }
    }
}
bool cmp(const struct road &a,const struct road &b)
{
    return a.costSum!=b.costSum?(a.costSum<b.costSum):(a.happinessSum!=b.happinessSum?(a.happinessSum>b.happinessSum):(a.happinessSum/(double)(a.path.size()-1)>b.happinessSum/(double)(b.path.size()-1)));
}
int main()
{
    int k,a,endCity;
    char c[15],s[15];
    scanf("%d%d%s",&n,&k,c);
    idToName.insert(make_pair(0,string(c)));
    nameToId.insert(make_pair(string(c),0));
    idToHappiness.insert(make_pair(0,0));
    for(int i=1; i<n; ++i)
    {
        scanf("%s%d",c,&a);
        idToName.insert(make_pair(i,string(c)));
        nameToId.insert(make_pair(string(c),i));
        idToHappiness.insert(make_pair(i,a));
        if(strcmp(c,"ROM")==0) endCity=i;
    }
    memset(mapx,0,sizeof(mapx));
    for(int i=0,ida,idb;i<k;++i)
    {
        scanf("%s%s%d",c,s,&a);
        ida=nameToId[string(c)];
        idb=nameToId[string(s)];
        mapx[ida][idb]=mapx[idb][ida]=a;
    }
    memset(visited,false,sizeof(visited));
    path.push_back(0);
    visited[0]=true;
    dfs(0,endCity,0,0);
    sort(ans.begin(),ans.end(),cmp);
    int out=ans[0].costSum,countx=1;
    for(;out==ans[countx].costSum;++countx);
    printf("%d %d %d %d\n",countx,ans[0].costSum,ans[0].happinessSum,(int)(ans[0].happinessSum/(double)(ans[0].path.size()-1)));
    printf("%s",idToName[ans[0].path[0]].c_str());
    for(int i=1;i<ans[0].path.size();++i)
         printf("->%s",idToName[ans[0].path[i]].c_str());
    printf("\n");
    return 0;
}