PAT（甲级）2019年秋季考试

第一题用搜索，超时了，待补

更新第一题思路
dfs + 剪枝，首先确定 n的最后一位数字肯定是9，为什么呢，因为 任意两个相邻的数肯定互为质数(gcd=1)，所以 n 的末尾肯定是9，这样n+1产生的各个位数和相加的和 才能有可能和sum(n)产生gcd>2的素数，
所以k可以剪枝成k-1，缩小一位，复杂度降低很多。。。这样估计就能AC了。
另外判断n和n+1两个数的各个位数之和的gcd是否为大于2的素数，可以提前预处理(数据量<90)，
第一题参考链接

7-1搜索 12/20分

原来写的代码，超时2个测试点

#include<bits/stdc++.h>
using namespace std;

int n,k,m;
typedef long long ll;
ll starts = 0;
ll endss = 0;
bool flag = false;
bool first = true;
int t = 1;

int sums(ll x){
    int ans = 0;
    while(x){
        ans += x%10;
        x/=10;
    }
    return ans;
}


ll gcd(ll a,ll b){
    if(b == 0) return a;
    return gcd(b,a%b);
}

bool solve(ll x){
    if(x <= 2) return false;
    for(int i=2;i<=sqrt(x);i++){
        if(x%i == 0) return false;
    }
    return true;
}


void dfs(int step,ll x){
    if(x >= endss) return;
    if(step > k+1) return;
    if(sums(x) > m) return;
    if(step == k+1) {
        int t1 = sums(x);
        if(sums(x) != m) return;
        int t2 = sums(x+1);
        ll g = gcd(t1,t2);
        if(solve(g)){
            if(first){
                first = false;
                cout<<"Case "<<t<<endl;
            }
            flag = true;
            cout<<t2<<" "<<x<<endl;
        }
        return;
    }
    for(int i=0;i<=9;i++){
        if(i==0 && step == 1) continue;
        ll temp = x * 10 + i;
        dfs(step+1,temp);
    }
} 

int main(){
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>k>>m;
        starts = pow(10,k-1);
        endss = pow(10,k);
        flag = false;   
        first = true;
        t = i;
        dfs(1,0);
        if(flag == false){
            cout<<"Case "<<t<<endl;
            cout<<"No Solution"<<endl;
        }
    }
    return 0;
}

7-2链表模拟 25

#include<bits/stdc++.h>
using namespace std;

const int maxn = 1e5+10;

struct node{
    int  address;
    int data;
    int  next;
}li[maxn];

int s1,s2;
int n; //不一定都是有效结点 
vector<node> v1;
vector<node> v2;
vector<node> ans;

int main(){
    scanf("%d%d%d",&s1,&s2,&n);
    for(int i=1;i<=n;i++){
        int dat,add,nex;
        scanf("%d%d%d",&add,&dat,&nex);
        li[add].address = add;
        li[add].data = dat;
        li[add].next = nex;
    }
    for(int cur = s1;cur != -1;cur = li[cur].next) v1.push_back(li[cur]);
    for(int cur = s2;cur != -1;cur = li[cur].next) v2.push_back(li[cur]);
    int len1 = v1.size() - 1;
    int len2 = v2.size() - 1;
    if(len1 >= 2*len2){
        reverse(v2.begin(),v2.end());
        int idx2 = 0;
        int idx1 = 0;
        while(idx2 <= len2){
            ans.push_back(v1[idx1++]);
            ans.push_back(v1[idx1++]);
            ans.push_back(v2[idx2++]);
        }
        while(idx1 <= len1) ans.push_back(v1[idx1++]);
    }else{
        reverse(v1.begin(),v1.end());
        int idx2 = 0;
        int idx1 = 0;
        while(idx1 <= len1){
            ans.push_back(v2[idx2++]);
            ans.push_back(v2[idx2++]);
            ans.push_back(v1[idx1++]);
        }
        while(idx2 <= len2) ans.push_back(v2[idx2++]);
    }
    n = ans.size();
    for(int i=0;i<n-1;i++){
        printf("%05d %d %05d\n",ans[i].address,ans[i].data,ans[i+1].address);
    }
    printf("%05d %d -1\n",ans[n-1].address,ans[n-1].data);
    return 0;
}

7-3树 25

#include<bits/stdc++.h>
using namespace std;

struct node{
    string dat;
    int lc,rc;
}tree[25];

int n;
int vis[25];

void dfs(int x){
    if(x > n) return;
    cout<<"(";
    if(tree[x].lc == -1 &&  tree[x].rc != -1){
        cout<<tree[x].dat;
        dfs(tree[x].rc);
        cout<<")";
    }else{
        if(tree[x].lc!=-1) dfs(tree[x].lc);
        if(tree[x].rc!=-1) dfs(tree[x].rc);
        cout<<tree[x].dat;
        cout<<")";
    }
}

int main(){
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>tree[i].dat>>tree[i].lc>>tree[i].rc;
        if(tree[i].lc != -1) vis[tree[i].lc] = 1;
        if(tree[i].rc != -1) vis[tree[i].rc] = 1;
    }
    int root = 1;
    for(int i=1;i<=n;i++){ //先确定 根的结点 没有当过 
        if(!vis[i]) root = i;
    }
    dfs(root);
    return 0;
}

7-4最短路 30

#include<bits/stdc++.h>
using namespace std;

const int inf = 0x3f3f3f3f;
const int maxn = 1e3+10;
const int maxm = 1e5+10;
struct edge{
    int v;
    int w;
    edge(int vv,int ww){
        v = vv;
        w = ww;
    }
};

vector<edge> g[maxn];
int n,e,k;
int di[maxn];
int dist[maxn];
int vis[maxn];

bool dijkstra(int s){
    memset(dist,inf,sizeof(dist));
    memset(vis,0,sizeof(vis));
    dist[s] = 0;
    for(int i=1;i<=n;i++){
        int v,min_w = inf;
        for(int j=1;j<=n;j++){
            if(!vis[j] && dist[j] < min_w){
                v = j;
                min_w = dist[j];
            }
        }
        if(min_w == inf) return false;
        vis[v] = 1;
        for(int j=0;j<g[v].size();j++){
            int x = g[v][j].v;
            int w = g[v][j].w;
            if(!vis[x] && dist[x] >dist[v] + w){
                dist[x] = dist[v] + w;
            }
        }
    }
    return true;
}
int vis2[maxn];
bool solve(){
    memset(vis2,0,sizeof(vis2));
    vis2[di[1]] = 1;
    for(int i=2;i<=n;i++){
        int mind = inf;
        for(int j=1;j<=n;j++){
            if(!vis2[j] && dist[j] <= mind){
                mind = dist[j];
            }
        }
        if(mind != dist[di[i]]) 
            return false;
        vis2[di[i]] = 1;
    }
    return true;
}


void solve2(){
    int visit[n + 1];
    for (int i = 1; i <= n; i++) visit[i] = 0;
    int start = di[1];
    visit[start] = 1;
    bool flag = false;
    for (int i = 2; i <= n; i++) {
        int mine = dist[di[i]];
        visit[di[i]] = 1;
        for (int j = 1; j <= n; j++) {
            if (visit[di[j]]) continue;
            int curv = dist[di[j]];
            if (mine >= curv) mine = curv;
        }
        if (mine < dist[di[i]]) {
            cout << "No" << endl;
            flag = true;
            break;
        }
    }
    if (flag == false) cout << "Yes" << endl;
}

int main(){
    cin>>n>>e;
    for(int i=1;i<=e;i++){
        int u,v,w;
        cin>>u>>v>>w;
        g[u].push_back(edge(v,w));
        g[v].push_back(edge(u,w));
    }
    cin>>k;
    for(int i=1;i<=k;i++){
        for(int j=1;j<=n;j++) cin>>di[j];
        int st = di[1];
        dijkstra(st);
//      solve2();
        if(solve()){
            cout<<"Yes"<<endl;
        }else{
            cout<<"No"<<endl;
        }
    }
    return 0;
}

来源：https://www.cnblogs.com/fisherss/p/11615671.html