PTA 天梯赛 L1

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=1,sum=0;
    for(r=1; ;r+=2){
    if(r==1)sum+=r;
    else sum+=r<<1;
    if(sum>n)break;
    }
    sum-=r<<1;
    r-=2;
//    cout<<sum<<" "<<r<<endl;    
    for(int i=1;i<=r;i++){
    for(int j=1;j<=r;j++){
    if(i<=j&&i+j<=r+1||i>=j&&i+j>=r+1)cout<<c;
    else if(i>j&&i+j<r+1) cout<<" ";
    
    }
    cout<<endl;
    }
    cout<<n-sum<<endl;
    return 0;    
}

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=0;i<len;i++){
    mp[s[i]-'0']++;
    }
    map<int,int>::iterator it;
    for(it=mp.begin();it!=mp.end();it++){
    cout<<it->fi<<":"<<it->se<<endl;
    }
    return 0;    
}

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=-1;
    for(int i=2;i<=sqrt(n)+1;i++){
    if(n%i==0)v.pb(i);
    }
    int len=v.size();
    for(int i=0;i<len;i++){
    int sum=v[i],cnt=0;
    for(int j=v[i]+1; ;j++){
    sum*=j;
    if(n%sum!=0)break;
    cnt++;
    }
    if(cnt>max){
    ans=v[i];
    max=cnt;
    }
    }
    if(len==0)cout<<1<<endl<<n<<endl;
    else {
    cout<<max+1<<endl;
    for(int i=0;i<=max;i++){
    if(i)cout<<"*";
    cout<<ans+i;    
    }
    cout<<endl;
    }
    return 0;    
}

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=0;
    int i,j;
    for(i=a,j=1;i<=b;j++,i++){
    printf("%5d",i);
    sum+=i;
    if(j%5==0)printf("\n");
    }
    if((j-1)%5!=0)printf("\n");
    printf("Sum = %d",sum);
//    cout<<"Sum = "<<sum<<endl;
    return 0;    

}

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==0?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=1;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>0&&b<0||a<0&&b<0)a=-a,b=-b;

    if(a>=b&&a%b!=0)printf("%lld %lld/%lld\n",a/b,a%b,b);
    else if(a/b!=0&&a%b==0)printf("%lld\n",a/b);
    else if(a/b==0&&a!=0)printf("%lld/%lld\n",a,b);
    else puts("0");

    return 0;    
}

 

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-9;
const ll MOD=1e9+9;
const int maxn=1e6+5;
bool flag[maxn];
int main(){
    int n,k,m,x;
    scanf("%d",&n);
    memset(flag,0,sizeof flag);
    while(n--){
    scanf("%d",&k);
    rep(i,1,k){
        scanf("%d",&x);
        if(k>1)flag[x]=1;
    }
    }
    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=0;i<v.size();i++)
    printf("%.5d%c",v[i],i==v.size()-1?'\n':' ');
    }
    return 0;
}

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-9;
const ll MOD=1e9+9;
const int maxn=1e6+5;
int f(string s){
    int num=0;
    int len=s.size();
    for(int i=0;i<len;i++)if(s[i]>'9'||s[i]<'0')return -1;
    for(int i=0;i<len;i++)num=num*10+s[i]-'0';
        return num;
}
int main(){
    string sa,sb;
    cin>>sa;
    getchar();
    getline(cin,sb);
    int a=f(sa);
    int b=f(sb);
    if(a<=1000&&a>=1&&(b<1||b>1000))
    cout<<a<<" + ? = ?";
    else if(b<=1000&&b>=1&&(a<1||a>1000))
    cout<<"? + "<<b<<" = ?";
    else if(a<=1000&&a>=1&&b<=1000&&b>=1)
    cout<<a<<" + "<<b<<" = "<<a+b<<endl;
    else cout<<"? + ? = ?"<<endl;
    return 0;
}

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-9;
const ll MOD=1e9+9;
const int maxn=1e6+5;
int main(){
    string s;
    cin>>s;
    set<int,greater<int> >st;
    int len=s.size();
    for(int i=0;i<len;i++){
    st.insert(s[i]-'0');
    }
    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=0;i<11;i++){
    for(int j=0;j<arr.size();j++){
        if(arr[j]==s[i]-'0')index.pb(j);
    }
    }
    cout<<"int[] arr = new int[]{";
    for(int i=0;i<arr.size();i++){
    if(i)cout<<",";
    cout<<arr[i];
    }
    cout<<"};"<<endl;
    cout<<"int[] index = new int[]{";
    for(int i=0;i<index.size();i++){
    if(i)cout<<",";
    cout<<index[i];
    }
    cout<<"};"<<endl;


    return 0;
}

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-9;
const ll MOD=1e9+9;
const int maxn=1e5+5;
struct node{int sex,vis;string name;};
struct ANS{string fi,se;};
int main(){
    int n;
    cin>>n;
    vector<node>v;
    rep(i,1,n){
    node tmp;
    cin>>tmp.sex>>tmp.name;
    tmp.vis=0;
    v.push_back(tmp);
    }
    vector<ANS>ans;
    int len=v.size();
    for(int i=0;i<len;i++){
    for(int j=len-1;j>=0;j--){
    if(v[i].sex!=v[j].sex&&v[j].vis==0&&v[i].vis==0){
    ANS tmp;
    tmp.fi=v[i].name;
    tmp.se=v[j].name;
    ans.pb(tmp);
    v[i].vis=1,v[j].vis=1;
    }
    }
    }
    len=ans.size();
    for(int i=0;i<len;i++)
        cout<<ans[i].fi<<" "<<ans[i].se<<endl;
    system("pause");
    return 0;

}

 

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-9;
const ll MOD=1e9+9;
const int maxn=1e5+5;
int main(){
    int n,num;
    cin>>n;
    string s;
    getchar();
    getline(cin,s);
    vector<char>v[200];
    int len=s.size();
    for(int i=len-1;i>=0;i--){
    if((i+1)%n==0)num=n;
    else num=(i+1)%n;
    v[num].pb(s[i]);
    }
    // cout<<"test :"<<endl;
    if(len%n==0)num=n;
    else num=len%n;
    for(int i=1;i<=n;i++){
    if(i>num)cout<<" ";
    for(int j=0;j<v[i].size();j++){
    cout<<v[i][j];
    }
    cout<<endl;
    }

    return 0;
}

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-9;
const ll MOD=1e9+9;
const int maxn=1e5+5;
int main(){
    int k;
    scanf("%d",&k);
    string s;
    int cnt=0,flag;
    while(cin>>s){
    if(s[0]=='E')break;
    cnt++;
    // cout<<"test "<<s<<" "<<cnt<<": "<<endl;
    if(cnt%(k+1)==0)flag=1;
    else flag=0;
    if(flag)cout<<s<<endl;
    else {
    if(s[0]=='C')cout<<"Bu"<<endl;
    else if(s[0]=='B')cout<<"JianDao"<<endl;
    else if(s[0]=='J')cout<<"ChuiZi"<<endl;
    }
    }
    // system("pause");
    return 0;
}

 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-9;
const ll MOD=1e9+9;
const int maxn=1e5+5;
int main(){
    ll ans,x,s=1,n=1;
    cin>>x;
    while(s<x)s=s*10+1,n++;
    while(1){
    cout<<s/x;
    s%=x;
    if(s==0){
    ans=s;
    break;
    }
    s=s*10+1;
    n++;
    }
    cout<<" "<<n<<endl;
    // system("pause");
    return 0;
}

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-9;
const ll MOD=1e9+9;
const int maxn=1e5+5;
char mp[200][200];
int main(){
    char c;
    int n;
    scanf("%c %d",&c,&n);
    rep(i,1,n){
    getchar();
    rep(j,1,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+1;
    bool issame=1;
    rep(i,1,n){
    rep(j,1,n){
        int x=cen-i;
        int y=cen-j;
        if(mp[x][y]!=mp[i][j])issame=0;
    }
    }
    if(issame){
    cout<<"bu yong dao le"<<endl;    
    rep(i,1,n){
    rep(j,1,n){
    if(mp[i][j]==' ')cout<<" ";
    else cout<<c;
    }
    cout<<endl;
    }
    return 0;
    }

    rep(i,1,n){
    rep(j,1,n){
    int x=cen-i;
    int y=cen-j;
    if(mp[x][y]!=' ')printf("%c",c);
    else printf(" ");
    }
    cout<<endl;
    }
// shit:    system("pause");
    return 0;
}

 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-9;
const ll MOD=1e9+9;
const int maxn=1e3+5;
struct book{bool flag;int time;};
int get(int h,int m){return h*60+m;}
book a[maxn];
int main(){
    rep(i,1,maxn)a[i].flag=0;
    int num,h,m,n;
    char op;
    scanf("%d",&n);
    while(n--){

    int cnt=0,time=0;
    while(1){
        scanf("%d %c %d:%d",&num,&op,&h,&m);
    if(num==0)break;
    if(op=='S'){
    a[num].flag=1,a[num].time=get(h,m);
    }
    else if(a[num].flag==1){
    time+=get(h,m)-a[num].time;
    a[num].flag=0;
    cnt++;
    }
    }
    if(cnt==0&&time==0)cout<<"0 0"<<endl;
    else printf("%d %.0lf\n",cnt,time*1.0/cnt);
    }
    // system("pause");
    return 0;
}

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-9;
const ll MOD=1e9+9;
const int maxn=1e3+5;
int team[maxn];
vector<int>ans[maxn];
int main(){

    int n,sum=0;
    scanf("%d",&n);
    rep(i,1,n){
    scanf("%d",&team[i]);
    team[i]*=10;
    }

    int pos=1,num=0;
    while(num<n){
    
    for(int i=1;i<=n;i++){
    if(team[i]!=0){
    team[i]--,ans[i].pb(pos);
    if(num==n-1)pos+=2;
    else pos++;
    if(team[i]==0)num++;
    }
    
    }

    }    
    rep(i,1,n){
    printf("#%d\n",i);
    for(int j=0;j<ans[i].size();j++){
    printf("%d%c",ans[i][j],(j+1)%10==0?'\n':' ');
    }
    }
    // system("pause");
    return 0;
}

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=0,flag2=0;
    int len=s.size();
    for(int i=0;i<len;i++){
    if(s[i]==','){
    if(s[i-1]=='g'&&s[i-2]=='n'&&s[i-3]=='o')flag1=1;
    }

    if(s[i]=='.'){
    if(s[i-1]=='g'&&s[i-2]=='n'&&s[i-3]=='o')flag2=1;
    }

    }
    if(flag1&&flag2){
    int cnt=0,seat;
    for(int i=len-1;i>=0;i--){
    if(s[i]==' ')cnt++;
    if(cnt>=3){
    seat=i;
    break;
    }
    }
    for(int i=0;i<=seat;i++)printf("%c",s[i]);
    cout<<"qiao ben zhong."<<endl;

    }

    else cout<<"Skipped"<<endl;

    }

    return 0;
}

 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-9;
const ll MOD=1e9+9;
const int maxn=1e3+5;
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=0;i<len;i++){
    if(s[i]=='?')s[i]='!';
    else if(isupper(s[i])&&s[i]!='I')s[i]=tolower(s[i]);
    } 
    for(int i=0;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=0;i<v.size();i++){
    if(v[i]=="I"||v[i]=="me")v[i]="you";
    else if(v[i]=="you"){
    if(i<2||!(v[i-1].size()==1&&!isalnum(v[i-1][0])))continue;
    if(v[i-2]=="can")v[i-2]="I",v[i]="can";
     if(v[i-2]=="could")v[i-2]="I",v[i]="could";
    }
    }
    for(int i=0;i<v.size();i++)cout<<v[i];
    cout<<endl;
    }
    system("pause");
    return 0;
}

 

未完待续；