PAT Advanced Level (1081~1087)

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;}