1136

Palindrome 回文
standard notation 标准符号
iteration 迭代

#include<string>
#include<iostream>
#include<algorithm>
using namespace std;

string plus0(string a, string b){
    int c = 0;
    string ans;
    for(int i = a.length()-1;i>=0;i--){
        char temp = c + a[i] + b[i] - '0';
        if(temp > '9'){
            temp -= 10;
            c = 1;
        }else{
            c = 0;
        }
        ans += temp;
    }
    if(c == 1){
        ans += '1';
    }
    reverse(ans.begin(), ans.end());
    return ans;
}

bool judge(string s){
    for(int i = 0, j = s.length()-1; i <= j; i++, j--){
        if(s[i]!=s[j]){
            return false;
        }
    }
    return true;
}

int main(){
    int i;
    string s1;
    cin>>s1;
    //别忘了首先判断s1 !!! 有两个测试用例是专门测试这个的！
    if(judge(s1)){ 
        cout<<s1<<" is a palindromic number."<<endl;
        return 0;
    }
    for(i=0;i<10;i++){
        string s2 = s1;
        reverse(s1.begin(), s1.end());
        string s3 = plus0(s1, s2);
        cout<<s2<<" + "<<s1<<" = "<<s3<<endl;
        if( judge(s3) ){
            cout<<s3<<" is a palindromic number."<<endl;
            break;
        }
        s1 = s3;
    } 
    if(i==10){
        printf("Not found in 10 iterations.\n");
    }
    return 0;
}

1137 STL

题目大意：

• 网上编程至少获得200分
• 期末考试至少60分（100分制）
• 按照公式计算最终得分

#include<stdio.h>
#include<iostream>
#include<string>
#include<map>
#include<vector>
#include<algorithm>
using namespace std;

typedef struct node{
    string id;
    int p, mid, final;
    int g;
}node;

bool cmp(node a, node b){
    if(a.g == b.g){
        return a.id < b.id;
    }
    return a.g > b.g;
}

int main(){
    int p, m, n;
    cin>>p>>m>>n;
    map<string, int> mp;
    vector<node> v;
    int count = 0;
    for(int i=0;i<p;i++){
        string id;
        int score;
        cin>>id>>score;
        if(score >= 200){
            count++;
            mp[id] = count;
            node temp;
            temp.id = id;
            temp.p = score;
            temp.mid = temp.final = temp.g = -1;
            v.push_back(temp);
        }
    }
    for(int i=0;i<m;i++){
        string id;
        int score;
        cin>>id>>score;
        if(mp[id]!=0){
            v[mp[id]-1].mid = score;
        }
    }
    for(int i=0;i<n;i++){
        string id;
        int score;
        cin>>id>>score;
        if(mp[id]!=0){
            v[mp[id]-1].final = score;
            v[mp[id]-1].g = v[mp[id]-1].mid > v[mp[id]-1].final ? 0.4*v[mp[id]-1].mid+0.6*v[mp[id]-1].final + 0.5 : v[mp[id]-1].final;
        }
    }
    sort(v.begin(), v.end(), cmp);
    for(int i=0;i<count && v[i].g >=60; i++){
        cout<<v[i].id<<" "<<v[i].p<<" "<<v[i].mid<<" "<<v[i].final<<" "<<v[i].g<<endl;
    }
    return 0;
}

1138 二叉树

题目大意： 给前序和中序遍历，输出后序遍历的第一个值 。

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

vector<int> preorder;
vector<int> inorder;
bool flag = false;
    
void find(int prest, int inst, int inen){
    if(inst > inen || flag == true){
        return;
    }
    int root = inst;
    while(inorder[root] != preorder[prest]){
        root++;
    }
    find(prest+1, inst, root-1);
    find(prest+root-inst+1, root+1, inen);
    if(flag == false){
        printf("%d", inorder[root]);
        flag = true;
    }
}

int main(){
    int n;
    cin>>n;
    preorder.resize(n);
    inorder.resize(n);
    for(int i=0;i<n;i++){
        cin>>preorder[i];
    }
    for(int i=0;i<n;i++){
        cin>>inorder[i];
    }
    find(0, 0, n-1);
    return 0;
}

1139

题目大意： A-C-D-B，知道A,B，找C 和D
analogously 类似的
注：

• 如果是同性恋，那么要保证a找的朋友不能是b，b找的朋友不能是a （3.4测试用例针对这一点，一开始我没考虑到 ）
• 考虑到0000之类的，还是用了string读入
• 第五个测试用例针对大数据，我就超时了555

#include<stdio.h>
#include<vector>
#include<algorithm>
#include<map>
#include<string>
#include<iostream>
using namespace std;

typedef struct node{
    string c,d;
}node;

map<string, bool> gender;//记录性别 
map<string, vector<string> > relation;//记录关系 
bool e[10000][10000];//记录是否存在关系 

bool cmp(node a, node b){
    if(a.c == b.c){
        return a.d < b.d;
    }
    return a.c < b.c;
}

int main(){
    int n,m;
    scanf("%d %d", &n, &m);
    for(int i=0;i<m;i++){
        string s1, s2;
        cin>>s1>>s2;
        if(s1.length() == 5){
            s1.erase(0,1);
            gender[s1] = true;
        }else{
            gender[s1] = false;
        }
        if(s2.length() == 5){
            s2.erase(0,1);
            gender[s2] = true;
        }else{
            gender[s2] = false;
        }
        relation[s1].push_back(s2); 
        relation[s2].push_back(s1);
        e[atoi(s1.c_str())][atoi(s2.c_str())] = e[atoi(s2.c_str())][atoi(s1.c_str())] = true;
    }
    
    int k;
    scanf("%d", &k);
    for(int i=0;i<k;i++){
        string s1, s2;
        cin>>s1>>s2;
        if(s1.length() == 5){
            s1.erase(0,1);
        }
        if(s2.length() == 5){
            s2.erase(0,1);
        }
        vector<node> ans;
        for(int j=0;j<relation[s1].size();j++){
            if(gender[relation[s1][j]] == gender[s1] && relation[s1][j] != s2){
                for(int p=0;p<relation[relation[s1][j]].size();p++){
                    if(gender[relation[relation[s1][j]][p]] == gender[s2] && relation[relation[s1][j]][p]!=s1){
                        if(e[ atoi(relation[relation[s1][j]][p].c_str()) ][ atoi(s2.c_str()) ] == true){
                            node temp;
                            temp.c = relation[s1][j];
                            temp.d = relation[relation[s1][j]][p];
                            ans.push_back(temp);
                        }
                    }
                } 
            }
        }
        sort(ans.begin(),ans.end(),cmp);
        printf("%d\n",ans.size());
        for(int j=0;j<ans.size();j++){
            cout<<ans[j].c<<" "<<ans[j].d<<endl;
        }
    }
    return 0;
}