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PAT（甲级）2017年秋季考试

还有一题由于上午心情复杂。。没调试完。待补。

A Cut Integer

模拟题

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

typedef long long ll;
int n;

int getLen(ll x){
    int len = 0;
    while(x){
        len++;
        x/=10;
    }
    return len;
}

int main(){
    cin>>n;
    while(n--){
        ll x;
        cin>>x;
        int len = getLen(x);
        ll temp = x;
        int t = 0;
        ll right = 0;
        ll rightTemp = 0;
        ll left = 0;
        while(t<len/2){
            rightTemp = rightTemp * 10 + temp%10;
            temp/=10;
            t++;
        }
        while(rightTemp){
            right = right * 10 + rightTemp%10;
            rightTemp/=10;
        }
        left = temp;
        if(left == 0 || right == 0){
            puts("No");
        }else{
            if(x%(left*right) == 0) puts("Yes");
            else puts("No");
        }
    }
    return 0;
} 

B Splitting A Linked List

链表题

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

const int maxn = 1010;
const int maxm = 100000;
struct node{
    int address;
    int data;
    int next;
}nod[maxm];

vector<node> vec;
vector<node> ans;
int Head,n,k;
int vis[maxm];

int main(){
    cin>>Head>>n>>k;
    for(int i=1;i<=n;i++){
        int add,dat,nex;
        cin>>add>>dat>>nex;
        nod[add].address = add;
        nod[add].data = dat;
        nod[add].next = nex;
    }
    for(int head = Head;head!=-1;head=nod[head].next){
        vec.push_back(nod[head]);
    }
    for(int i=0;i<vec.size();i++){
        if(vec[i].data < 0 && !vis[vec[i].address]){
            ans.push_back(vec[i]);
            vis[vec[i].address] = 1;
        }
    }
    for(int i=0;i<vec.size();i++){
        if(vec[i].data >= 0 && vec[i].data <= k && !vis[vec[i].address]){
            ans.push_back(vec[i]);
            vis[vec[i].address] = 1;
        }
    }
    for(int i=0;i<vec.size();i++){
        if(!vis[vec[i].address]){
            ans.push_back(vec[i]);
            vis[vec[i].address] = 1;
        }
    }
    if(ans.size() == 1){
        printf("%05d %d -1",ans[0].address,ans[0].data);
        return 0;
    }
    for(int i=0;i<ans.size()-1;i++){
        printf("%05d %d %05d\n",ans[i].address,ans[i].data,ans[i+1].address);
    }
    if(ans.size() > 1) printf("%05d %d -1",ans[ans.size()-1].address,ans[ans.size()-1].data);
    return 0;
}

C Vertex Cover

简单图论，最小覆盖，邻接表存图

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

const int maxn = 10000;
int n,m,k;
int g[maxn][maxn];
int vis[maxn][maxn];
vector<int> vec;
void init(){
    for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            vis[i][j] = 0;
        }
    }
}

int main(){
    cin>>n>>m;
    for(int i=0;i<m;i++){
        int u,v;
        cin>>u>>v;
        g[u][v] = 1;
        g[v][u] = 1;
    }
    cin>>k;
    for(int t=0;t<k;t++){
        init();
        int nv;
        cin>>nv;
        for(int i=0;i<nv;i++) {
            int d;
            cin>>d;
            vec.push_back(d);
        }
        for(int i=0;i<=nv-1;i++){
            for(int j=0;j<n;j++){
                if(g[vec[i]][j] == 1){
                    vis[vec[i]][j] = 1;
                    vis[j][vec[i]] = 1;
                }
            }
        }
        bool flag = true;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(g[i][j] == 1 && (vis[i][j] == 0 || vis[j][i] == 0)){
                    flag = false;
                    break;
                }
            }
            if(flag == false) break;
        }
        if(flag) puts("Yes");
        else puts("No");
        vec.clear();
    }
    return 0;
}

D Is It A Red-Black Tree

判断是否红黑树
待补

PAT（甲级）2017年秋季考试

标签：||   mes   const   size   tle   复杂   flag   main   ESS   

原文地址：https://www.cnblogs.com/fisherss/p/11988440.html