技术教程 编程教程 PTA 天梯赛 L1 正文

PTA 天梯赛 L1

2023-04-20,,

L1-002 打印沙漏

细节：就是在  (i>j&&i+j<r+1) 这个区间里才有空格，然后就是 for 循环   for(r=1; ;r+=2)  条件不满足之后还会再+2；

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

typedef long double ldb;

int main(){

    int n;

    char c;

    scanf("%d %c",&n,&c);

//    int r=1,sum=1;

//    for(r=3;sum<=n ;r+=2){

//    sum+=r*2;

//    }

//    sum-=r*2;

//    r-=2;

    int r=,sum=;

    for(r=; ;r+=){

    if(r==)sum+=r;

    else sum+=r<<;

    if(sum>n)break;

    }

    sum-=r<<;

    r-=;

//    cout<<sum<<" "<<r<<endl;

    for(int i=;i<=r;i++){

    for(int j=;j<=r;j++){

    if(i<=j&&i+j<=r+||i>=j&&i+j>=r+)cout<<c;

    else if(i>j&&i+j<r+) cout<<" ";

    }

    cout<<endl;

    }

    cout<<n-sum<<endl;

    return ;

}

L1-003

搞个map即可，然后注意map只能从first  映射到second

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

typedef long double ldb;

int main(){

    string s;

    cin>>s;

    map<int,int>mp;

    int len=s.size();

    for(int i=;i<len;i++){

    mp[s[i]-'']++;

    }

    map<int,int>::iterator it;

    for(it=mp.begin();it!=mp.end();it++){

    cout<<it->fi<<":"<<it->se<<endl;

    }

    return ;

}

L1-006 连续因子

解法：就是你把他的全部因子求出来，然后去枚举每一个因子的最大连续次数，最后取一个最大值即可；

然后如果n是素数，输出n即可；

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

typedef long double ldb;

int main(){

    int n;

    cin>>n;

    vector<int>v;

    int ans,max=-;

    for(int i=;i<=sqrt(n)+;i++){

    if(n%i==)v.pb(i);

    }

    int len=v.size();

    for(int i=;i<len;i++){

    int sum=v[i],cnt=;

    for(int j=v[i]+; ;j++){

    sum*=j;

    if(n%sum!=)break;

    cnt++;

    }

    if(cnt>max){

    ans=v[i];

    max=cnt;

    }

    }

    if(len==)cout<<<<endl<<n<<endl;

    else {

    cout<<max+<<endl;

    for(int i=;i<=max;i++){

    if(i)cout<<"*";

    cout<<ans+i;

    }

    cout<<endl;

    }

    return ;

}

L1-008 求整数段和

细节，最后不要打出两个换行即可；

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

typedef long double ldb;

int main(){

    int a,b;

    cin>>a>>b;

    ll sum=;

    int i,j;

    for(i=a,j=;i<=b;j++,i++){

    printf("%5d",i);

    sum+=i;

    if(j%==)printf("\n");

    }

    if((j-)%!=)printf("\n");

    printf("Sum = %d",sum);

//    cout<<"Sum = "<<sum<<endl;

    return ;    

}

L1-009 N个数求和

细节：这里就是模拟通分，然后注意每一步都要约分，不然爆long long，然后这里我犯一些细节错误，就是

判断语句应该这样写： if(a/b!=0&&a%b!=0)

而不是这样：

if(a>=b&&a%b!=0）

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

ll gcd(ll a,ll b){return b==?a:gcd(b,a%b);}

int main(){

    int n;

    scanf("%d",&n);

    ll g,a,b,c,d;

    scanf("%lld/%lld",&a,&b);

    g=gcd(a,b);

    a/=g;

    b/=g;

    for(int i=;i<n;i++){

    scanf("%lld/%lld",&c,&d);

    ll tmp=gcd(b,d);

    a=a*d/tmp+c*b/tmp;

    b=b/gcd(b,d)*d;

    g=gcd(a,b);

    a/=g;

    b/=g;

    }

    if(a>&&b<||a<&&b<)a=-a,b=-b;

    if(a>=b&&a%b!=)printf("%lld %lld/%lld\n",a/b,a%b,b);

    else if(a/b!=&&a%b==)printf("%lld\n",a/b);

    else if(a/b==&&a!=)printf("%lld/%lld\n",a,b);

    else puts("");

    return ;

}

就是考虑负数的情况这样就不对了；

L1-020 帅到没朋友

注意是五位ID就行了，然后搞个set判重，vector保存结果；

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define pf push_front

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

typedef long double ldb;

typedef double db;

const db PI=acos(-1.0);

const ll INF=0x3f3f3f3f3f3f3f3fLL;

const int inf=0x3f3f3f3f;//0x7fffffff;

const double eps=1e-;

const ll MOD=1e9+;

const int maxn=1e6+;

bool flag[maxn];

int main(){

    int n,k,m,x;

    scanf("%d",&n);

    memset(flag,,sizeof flag);

    while(n--){

    scanf("%d",&k);

    rep(i,,k){

        scanf("%d",&x);

        if(k>)flag[x]=;

    }

    }

    set<int>st;

    scanf("%d",&m);

    vector<int>v;

    while(m--){

        scanf("%d",&x);

        if(!flag[x]&&st.find(x)==st.end())v.pb(x);

        if(!flag[x])st.insert(x);

    }

    if(st.empty())cout<<"No one is handsome";

    else {

    for(int i=;i<v.size();i++)

    printf("%.5d%c",v[i],i==v.size()-?'\n':' ');

    }

    return ;

}

L1-025 正整数A+B

就是写个函数操作会很方便，要学会结构化程序设计；

然后注意读入的时候第二个要用getline读入

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define pf push_front

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

typedef long double ldb;

typedef double db;

const db PI=acos(-1.0);

const ll INF=0x3f3f3f3f3f3f3f3fLL;

const int inf=0x3f3f3f3f;//0x7fffffff;

const double eps=1e-;

const ll MOD=1e9+;

const int maxn=1e6+;

int f(string s){

    int num=;

    int len=s.size();

    for(int i=;i<len;i++)if(s[i]>''||s[i]<'')return -;

    for(int i=;i<len;i++)num=num*+s[i]-'';

        return num;

}

int main(){

    string sa,sb;

    cin>>sa;

    getchar();

    getline(cin,sb);

    int a=f(sa);

    int b=f(sb);

    if(a<=&&a>=&&(b<||b>))

    cout<<a<<" + ? = ?";

    else if(b<=&&b>=&&(a<||a>))

    cout<<"? + "<<b<<" = ?";

    else if(a<=&&a>=&&b<=&&b>=)

    cout<<a<<" + "<<b<<" = "<<a+b<<endl;

    else cout<<"? + ? = ?"<<endl;

    return ;

}

L1-027 出租

注意 自定义的set这样写：set<int,greater<int> >st;

搞个vector就行了；

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define pf push_front

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

typedef long double ldb;

typedef double db;

const db PI=acos(-1.0);

const ll INF=0x3f3f3f3f3f3f3f3fLL;

const int inf=0x3f3f3f3f;//0x7fffffff;

const double eps=1e-;

const ll MOD=1e9+;

const int maxn=1e6+;

int main(){

    string s;

    cin>>s;

    set<int,greater<int> >st;

    int len=s.size();

    for(int i=;i<len;i++){

    st.insert(s[i]-'');

    }

    vector<int>arr;

    set<int,greater<int> >::iterator it;

    for(it=st.begin();it!=st.end();it++){

        int x=*it;

    arr.pb(x);

    }

    vector<int>index;

    for(int i=;i<;i++){

    for(int j=;j<arr.size();j++){

        if(arr[j]==s[i]-'')index.pb(j);

    }

    }

    cout<<"int[] arr = new int[]{";

    for(int i=;i<arr.size();i++){

    if(i)cout<<",";

    cout<<arr[i];

    }

    cout<<"};"<<endl;

    cout<<"int[] index = new int[]{";

    for(int i=;i<index.size();i++){

    if(i)cout<<",";

    cout<<index[i];

    }

    cout<<"};"<<endl;

    return ;

}

L1-030 一帮一

就是vector  你不能这样读进来

错误写法：

rep(i,1,n)scanf("%d",&v[i]);

正确写法：

rep(i,1,n)scanf("%d",&x),v.pb(x);

不会用pair，只会结构，这题可能用pair更好做吧；

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define pf push_front

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

typedef long double ldb;

typedef double db;

const db PI=acos(-1.0);

const ll INF=0x3f3f3f3f3f3f3f3fLL;

const int inf=0x3f3f3f3f;//0x7fffffff;

const double eps=1e-;

const ll MOD=1e9+;

const int maxn=1e5+;

struct node{int sex,vis;string name;};

struct ANS{string fi,se;};

int main(){

    int n;

    cin>>n;

    vector<node>v;

    rep(i,,n){

    node tmp;

    cin>>tmp.sex>>tmp.name;

    tmp.vis=;

    v.push_back(tmp);

    }

    vector<ANS>ans;

    int len=v.size();

    for(int i=;i<len;i++){

    for(int j=len-;j>=;j--){

    if(v[i].sex!=v[j].sex&&v[j].vis==&&v[i].vis==){

    ANS tmp;

    tmp.fi=v[i].name;

    tmp.se=v[j].name;

    ans.pb(tmp);

    v[i].vis=,v[j].vis=;

    }

    }

    }

    len=ans.size();

    for(int i=;i<len;i++)

        cout<<ans[i].fi<<" "<<ans[i].se<<endl;

    system("pause");

    return ;

}

L1-039 古风排版

这题挺有意思的；也不是很难；

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define pf push_front

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

typedef long double ldb;

typedef double db;

const db PI=acos(-1.0);

const ll INF=0x3f3f3f3f3f3f3f3fLL;

const int inf=0x3f3f3f3f;//0x7fffffff;

const double eps=1e-;

const ll MOD=1e9+;

const int maxn=1e5+;

int main(){

    int n,num;

    cin>>n;

    string s;

    getchar();

    getline(cin,s);

    vector<char>v[];

    int len=s.size();

    for(int i=len-;i>=;i--){

    if((i+)%n==)num=n;

    else num=(i+)%n;

    v[num].pb(s[i]);

    }

    // cout<<"test :"<<endl;

    if(len%n==)num=n;

    else num=len%n;

    for(int i=;i<=n;i++){

    if(i>num)cout<<" ";

    for(int j=;j<v[i].size();j++){

    cout<<v[i][j];

    }

    cout<<endl;

    }

    return ;

}

L1-044 稳赢

注意取余这么写：cnt%(k+1)==0

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define pf push_front

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

typedef long double ldb;

typedef double db;

const db PI=acos(-1.0);

const ll INF=0x3f3f3f3f3f3f3f3fLL;

const int inf=0x3f3f3f3f;//0x7fffffff;

const double eps=1e-;

const ll MOD=1e9+;

const int maxn=1e5+;

int main(){

    int k;

    scanf("%d",&k);

    string s;

    int cnt=,flag;

    while(cin>>s){

    if(s[]=='E')break;

    cnt++;

    // cout<<"test "<<s<<" "<<cnt<<": "<<endl;

    if(cnt%(k+)==)flag=;

    else flag=;

    if(flag)cout<<s<<endl;

    else {

    if(s[]=='C')cout<<"Bu"<<endl;

    else if(s[]=='B')cout<<"JianDao"<<endl;

    else if(s[]=='J')cout<<"ChuiZi"<<endl;

    }

    }

    // system("pause");

    return ;

}

L1-046 整除光棍

这题本来想暴力，但必然是妥妥的T了

然后考虑模拟手算的除法；

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define pf push_front

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

typedef long double ldb;

typedef double db;

const db PI=acos(-1.0);

const ll INF=0x3f3f3f3f3f3f3f3fLL;

const int inf=0x3f3f3f3f;//0x7fffffff;

const double eps=1e-;

const ll MOD=1e9+;

const int maxn=1e5+;

int main(){

    ll ans,x,s=,n=;

    cin>>x;

    while(s<x)s=s*+,n++;

    while(){

    cout<<s/x;

    s%=x;

    if(s==){

    ans=s;

    break;

    }

    s=s*+;

    n++;

    }

    cout<<" "<<n<<endl;

    // system("pause");

    return ;

}

L1-054 福到了

这题推了好久，其实就一个东西，旋转180度的话，要交换的满足 x1+x2=n+1,y1+y2=n+1;

因为他们绕着中心旋转180；

也不用转换，直接判断输出就是了；

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define pf push_front

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

typedef long double ldb;

typedef double db;

const db PI=acos(-1.0);

const ll INF=0x3f3f3f3f3f3f3f3fLL;

const int inf=0x3f3f3f3f;//0x7fffffff;

const double eps=1e-;

const ll MOD=1e9+;

const int maxn=1e5+;

char mp[][];

int main(){

    char c;

    int n;

    scanf("%c %d",&c,&n);

    rep(i,,n){

    getchar();

    rep(j,,n){

    scanf("%c",&mp[i][j]);

    }

    }

    // cout<<"test :"<<endl;

    // rep(i,1,n){

    // rep(j,1,n){

    // printf("%c",mp[i][j]);

    // }

    // cout<<endl;

    // }

    int cen=n+;

    bool issame=;

    rep(i,,n){

    rep(j,,n){

        int x=cen-i;

        int y=cen-j;

        if(mp[x][y]!=mp[i][j])issame=;

    }

    }

    if(issame){

    cout<<"bu yong dao le"<<endl;

    rep(i,,n){

    rep(j,,n){

    if(mp[i][j]==' ')cout<<" ";

    else cout<<c;

    }

    cout<<endl;

    }

    return ;

    }

    rep(i,,n){

    rep(j,,n){

    int x=cen-i;

    int y=cen-j;

    if(mp[x][y]!=' ')printf("%c",c);

    else printf(" ");

    }

    cout<<endl;

    }

// shit:    system("pause");

    return ;

}

L1-043 阅览室 :

其实就是遍历，注意一下细节就行了；

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define pf push_front

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

typedef long double ldb;

typedef double db;

const db PI=acos(-1.0);

const ll INF=0x3f3f3f3f3f3f3f3fLL;

const int inf=0x3f3f3f3f;//0x7fffffff;

const double eps=1e-;

const ll MOD=1e9+;

const int maxn=1e3+;

struct book{bool flag;int time;};

int get(int h,int m){return h*+m;}

book a[maxn];

int main(){

    rep(i,,maxn)a[i].flag=;

    int num,h,m,n;

    char op;

    scanf("%d",&n);

    while(n--){

    int cnt=,time=;

    while(){

        scanf("%d %c %d:%d",&num,&op,&h,&m);

    if(num==)break;

    if(op=='S'){

    a[num].flag=,a[num].time=get(h,m);

    }

    else if(a[num].flag==){

    time+=get(h,m)-a[num].time;

    a[num].flag=;

    cnt++;

    }

    }

    if(cnt==&&time==)cout<<"0 0"<<endl;

    else printf("%d %.0lf\n",cnt,time*1.0/cnt);

    }

    // system("pause");

    return ;

}

L1-049 天梯赛座位分配

挺有意思的，注意格式输出，然后就是用一个nun记录已经安排满的队伍，如果没满就一直放，再记个pos表示位置，就行了；

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define pf push_front

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

typedef long double ldb;

typedef double db;

const db PI=acos(-1.0);

const ll INF=0x3f3f3f3f3f3f3f3fLL;

const int inf=0x3f3f3f3f;//0x7fffffff;

const double eps=1e-;

const ll MOD=1e9+;

const int maxn=1e3+;

int team[maxn];

vector<int>ans[maxn];

int main(){

    int n,sum=;

    scanf("%d",&n);

    rep(i,,n){

    scanf("%d",&team[i]);

    team[i]*=;

    }

    int pos=,num=;

    while(num<n){

    for(int i=;i<=n;i++){

    if(team[i]!=){

    team[i]--,ans[i].pb(pos);

    if(num==n-)pos+=;

    else pos++;

    if(team[i]==)num++;

    }

    }

    }

    rep(i,,n){

    printf("#%d\n",i);

    for(int j=;j<ans[i].size();j++){

    printf("%d%c",ans[i][j],(j+)%==?'\n':' ');

    }

    }

    // system("pause");

    return ;

}

L1-059 敲笨钟

做法：暴力枚举一下，然后注意读掉回车；

#include<bits/stdc++.h>

using namespace std;

#define per(i,j,k) for(int i=k;i>=j;i--)

int main(){

    int n;

    cin>>n;

    string s;

    getchar();

    while(n--){

    getline(cin,s);

    int flag1=,flag2=;

    int len=s.size();

    for(int i=;i<len;i++){

    if(s[i]==','){

    if(s[i-]=='g'&&s[i-]=='n'&&s[i-]=='o')flag1=;

    }

    if(s[i]=='.'){

    if(s[i-]=='g'&&s[i-]=='n'&&s[i-]=='o')flag2=;

    }

    }

    if(flag1&&flag2){

    int cnt=,seat;

    for(int i=len-;i>=;i--){

    if(s[i]==' ')cnt++;

    if(cnt>=){

    seat=i;

    break;

    }

    }

    for(int i=;i<=seat;i++)printf("%c",s[i]);

    cout<<"qiao ben zhong."<<endl;

    }

    else cout<<"Skipped"<<endl;

    }

    return ;

}

L1-064 估值一亿的AI核心代码

这题挺有意思的；

解法：如果给你一个字符串，单纯让你替换单词，频繁用string的搜索是很麻烦的，参考了大神的做法；

就是搞个vector<string>v ; 这样你每次截取单词，空格，和标点符号就很方便，然后按题目意思模拟即可；

PS：有个函数 isalnum 就是判断是否为数字或者字母；

#include<bits/stdc++.h>

using namespace std;

#define rep(i,j,k) for(int i=(int)j;i<=(int)k;i++)

#define per(i,j,k) for(int i=(int)k;i>=(int)j;i--)

#define pb push_back

#define pf push_front

#define fi first

#define se second

typedef long long ll;

typedef unsigned long long ull;

typedef long double ldb;

typedef double db;

const db PI=acos(-1.0);

const ll INF=0x3f3f3f3f3f3f3f3fLL;

const int inf=0x3f3f3f3f;//0x7fffffff;

const double eps=1e-;

const ll MOD=1e9+;

const int maxn=1e3+;

int main(){

     int n;

     cin>>n;

     getchar();

    while(n--){

    string s;

    vector<string>v;

    getline(cin,s);

    cout<<s<<endl<<"AI: ";

    int len=s.size();

    for(int i=;i<len;i++){

    if(s[i]=='?')s[i]='!';

    else if(isupper(s[i])&&s[i]!='I')s[i]=tolower(s[i]);

    }

    for(int i=;i<len; ){

    string tmp="";

    if(isalpha(s[i])){

    while(isalpha(s[i])&&i<len)tmp+=s[i++];

    }

    else if(isdigit(s[i])){

    while(isdigit(s[i])&&i<len)tmp+=s[i++];

    }

    else if(s[i]==' '){

    tmp=" ";

    while(s[i]==' ')i++;

    }

    else {

    tmp+=s[i++];

    if(!v.empty()&&v.back()==" ")v.pop_back();

    }

    if(tmp==" "&&(v.empty()||i==len))continue;

    v.pb(tmp);

    }

    for(int i=;i<v.size();i++){

    if(v[i]=="I"||v[i]=="me")v[i]="you";

    else if(v[i]=="you"){

    if(i<||!(v[i-].size()==&&!isalnum(v[i-][])))continue;

    if(v[i-]=="can")v[i-]="I",v[i]="can";

     if(v[i-]=="could")v[i-]="I",v[i]="could";

    }

    }

    for(int i=;i<v.size();i++)cout<<v[i];

    cout<<endl;

    }

    system("pause");

    return ;

}

未完待续；

PTA 天梯赛 L1的相关教程结束。

《PTA 天梯赛 L1.doc》

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