PAT甲级2019秋季考试题解

7-1 Forever 

判断区间内的符合条件的数并输出

直接暴力加排序会有两个点tle，加点剪枝卡测试点就ac，具体看代码吧，近几年的a题都是这个风格~

#include<bits/stdc++.h>
using namespace std;
const int maxn=1014;
struct node {
    int data;
    int n;
};
int gcd (int a,int b) {
    return b==0?a:gcd(b,a%b);
}
int isprime (int n) {
    if (n<=2) return 0;
    if (n%2==0) return 0;
    for (int i=3;i*i<=n;i++)
    if (n%i==0) return 0;
    return 1;
}
int digitsum (int n) {
    string s=to_string(n);
    int ans=0;
    for (int i=0;i<s.length();i++)
    ans+=s[i]-'0';
    return ans;
}
bool cmp (node a,node b) {
    if (a.n!=b.n) return a.n<b.n;
    else return a.data<b.data;
}
int main () {
    int N,K,M;
    scanf ("%d",&N);
    for (int i=0;i<N;i++) {
        scanf ("%d %d",&K,&M);
        int flag=0;
        printf ("Case %d\n",i+1);
        //if (M>=9*K) {
            //printf ("No Solution\n");
            //continue;
       // }
       vector<node> vi;
        for (int j=pow(10,K-1)+pow(10,K-2)-1;j<pow(10,K);j+=100) {
            //if (j%10!=9) continue;
            if (digitsum(j)!=M) continue;
            int n=digitsum(j+1);
            if (isprime(gcd(M,n))) {
                //flag++;
                vi.push_back({j,digitsum(j+1)});
                //printf ("%d %d\n",n,j);
            }
        }
        if (vi.size()==0) printf ("No Solution\n");
        else {
            sort (vi.begin(),vi.end(),cmp);
            for (int j=0;j<vi.size();j++) {
                printf ("%d %d\n",vi[j].n,vi[j].data);
            }
        }
    }
    return 0;
}

7-2 Merging Linked Lists

链表重排，用结构体实现静态链表并输出，老模板题了~

#include<bits/stdc++.h>
using namespace std;
const int maxn=1014;
struct node {
    int address;
    int data;
    int next;
}Node[100010];
int main () {
    int head1,head2,n;
    scanf ("%d %d %d",&head1,&head2,&n);
    int address,data,next;
    for (int i=0;i<n;i++) {
        scanf ("%d %d %d",&address,&data,&next);
        Node[address].address=address;
        Node[address].data=data;
        Node[address].next=next;
    }
    vector<node> l1,l2;
    while (head1!=-1) {
        l1.push_back(Node[head1]);
        head1=Node[head1].next;
    }
    while (head2!=-1) {
        l2.push_back(Node[head2]);
        head2=Node[head2].next;
    }
    int len1=l1.size(),len2=l2.size(),cnt=0;
    if (len1>len2) {
    printf ("%05d %d ",l1[0].address,l1[0].data);
    for (int i=1;i<len1;i++) {
        printf ("%05d\n%05d %d ",l1[i].address,l1[i].address,l1[i].data);
        if ((i+1)%2==0&&cnt<len2) {
            printf ("%05d\n%05d %d ",l2[len2-1-cnt].address,l2[len2-1-cnt].address,l2[len2-1-cnt].data);
            cnt++;
        }
        if (i==len1-1) {
            printf ("-1");
        }
    }
    }
    else {
    printf ("%05d %d ",l2[0].address,l2[0].data);
    for (int i=1;i<len2;i++) {
        printf ("%05d\n%05d %d ",l2[i].address,l2[i].address,l2[i].data);
        if ((i+1)%2==0&&cnt<len1) {
            printf ("%05d\n%05d %d ",l1[len1-1-cnt].address,l1[len1-1-cnt].address,l1[len1-1-cnt].data);
            cnt++;
        }
        if (i==len2-1) {
            printf ("-1");
        }
    }
    }
    return 0;
}

7-3 Postfix Expression

对表达式树进行dfs，对着样例瞎搞搞就ac，具体看代码~

#include <iostream>
#include <string>

using namespace std;

struct node {
    string s;
    int lChild, rChild, parent = -1;
};

node nodes[30];

void dfs(int root) {
    if ((nodes[root].lChild == -1 || nodes[root].rChild == -1) && nodes[root].lChild != nodes[root].rChild) {
        cout << "(" << nodes[root].s;
        if (nodes[root].lChild != -1) dfs(nodes[root].lChild);
        if (nodes[root].rChild != -1) dfs(nodes[root].rChild);
        cout << ")";
    } else if (nodes[root].lChild != -1) {
        cout << "(";
        dfs(nodes[root].lChild);
        dfs(nodes[root].rChild);
        cout << nodes[root].s << ")";
    } else {
        cout << "(" << nodes[root].s << ")";
    }
}

int main() {
    int n;
    cin >> n;

    for (int i = 1; i <= n; ++i) {
        cin >> nodes[i].s >> nodes[i].lChild >> nodes[i].rChild;
        nodes[nodes[i].lChild].parent = i;
        nodes[nodes[i].rChild].parent = i;
    }

    int root = 1;
    while (nodes[root].parent != -1) root = nodes[root].parent;

    dfs(root);
}

 

7-4 Dijkstra Sequence

用floyd算法算出起点到所有结点的最短距离，再判断给定序列里的点到起点的最短距离是否是升序~

#include<bits/stdc++.h>
using namespace std;
const int maxn=1014;
const int inf=1e9;
int d[maxn][maxn];
int mind[maxn];
int N,M;
void floyd () {
    for (int k=1;k<=N;k++)
    for (int i=1;i<=N;i++)
    for (int j=1;j<=N;j++)
    if (i!=j) {
        d[i][j]=min(d[i][j],d[i][k]+d[k][j]);
    }
}
int main () {
    scanf ("%d %d",&N,&M);
    for (int i=1;i<=N;i++)
    for (int j=1;j<=N;j++)
    d[i][j]=inf;
    int u,v;
    for (int i=0;i<M;i++) {
        scanf ("%d %d",&u,&v);
        scanf ("%d",&d[u][v]);
        d[v][u]=d[u][v];
    }
    floyd ();
    int k;
    scanf ("%d",&k);
    int a[maxn];
    for (int i=0;i<k;i++) {
        scanf ("%d",&a[0]);
        int flag=0;
        for (int j=1;j<N;j++) {
            scanf ("%d",&a[j]);
        }
        for (int j=1;j<N-1;j++) {
            if (d[a[0]][a[j]]>d[a[0]][a[j+1]]) flag++;
        }
        if (flag==0) printf ("Yes\n");
        else printf ("No\n");
    }
    return 0;
}