pat (basic level)题解

想要锻炼自己的代码能力，看自己能否用最简练的代码AC。下面开始：

1006

简单的字符转换没什么难度

#include<iostream>
using namespace std;
int main(){
    int n;
    cin >> n;
    int bai,shi,ge;
    bai = n / 100;
    shi = (n - bai * 100) / 10;
    ge = n % 10;
    for(int i = 0; i < bai; i ++){
        cout << 'B';
    }
    for(int i = 0; i < shi; i++){
        cout << 'S';
    }
    for(int i = 0; i < ge; i++){
        cout << i + 1;
    }
    return 0;
}

1007

这里主要是不要偷懒质数一定要循环到sqrt(n)才好，其次是要判定i+2有没有越界这是个坑要注意。

#include<iostream>
#include<math.h>
using namespace std;
bool check(int n){
    bool flag = true;
    for(int i = 2; i <= sqrt(n); i++){
        if(n % i == 0){
            flag = false;
            break;
        }
    }
    return flag;
}
int main(){
    int n;
    cin >> n;
    int sum = 0;
    for(int i = 2; i < n; i++){
        if(check(i)){
            if(check(i + 2) && (i + 2) <= n) ++sum;
        } 
    }
    cout << sum;
    return 0;
}

1008逻辑移位，注意取模

#include<iostream>
using namespace std;
int main(){
    int n, m;
    cin >>n >>m;
    int a[n];
    for(int i = 0; i < n; i++){
        cin >> a[i];
    }
    int t = (-m % n) + n;
    for(int i = 0; i < n; i++){
        cout << a[((t + i) % n)];
        if(i == n - 1) continue;
        cout << " ";
    }
    return 0;
}

 1009 字符串处理，注意单行读入

#include<iostream>
#include<string>
using namespace std;
int main(){
    string s;
    string t;
    getline(cin,s);
    t = s;
    int length = t.length();
    while(t.find(' ') >= 0 && t.find(' ') < length){
        int pos = t.rfind(' ');
        cout << t.substr(pos + 1) << ' ';
        t = t.substr(0, pos);
        length = pos - 1;    
    }
    cout << t;
    return 0;
}

1010 难点在于无结束标志的数字读入

scanf("%d%d", &n, &e) != EOF

#include<stdio.h>
using namespace std;
int main(){
    int n, e, flag = 0;
    while(scanf("%d%d", &n, &e) != EOF){
        if(n * e){
            if(flag){
                printf(" ");
            }
            else flag = 1;
            printf("%d %d", n * e, e - 1);
        }
    }
    if(!flag) printf("0 0");
  return 0;  
}

1011 此题涉及大数字运算可以用long long来计算

 

#include<iostream>
#include<stdio.h>
using namespace std;
int main(){
    int n;
    long long a, b, c;
    cin >> n;
    for(int i = 0; i < n; i++){
        cin >> a >> b >> c;
        if(a + b > c) printf("Case #%d: true\n", i + 1);
        else printf("Case #%d: false\n", i + 1);
    }
    return 0;
    
}

 1012 简单的判断，考察浮点数的保留小数位数

#include<iostream>
#include <iomanip>
using namespace std;
int main(){
    int n;
    cin >> n;
    int x;
    int a[6] ={0};
    double a4sum = 0;
    int sign = 1;
    bool flag[6] = {true,true,true,true,true,true};
    while(n--){
        cin >> x;    
        if(x % 10 == 0) {
            a[1] += x;
            flag[1] = false;
        }
        if(x % 5 == 1) {
            a[2] += x * sign;
            sign = -sign;
            flag[2] = false;
        }
        if(x % 5 == 2){
            flag[3] = false;
            ++a[3];
        } 
        if(x % 5 == 3){
            ++a[4];
            a4sum += x;
            flag[4] = false;
        }
        if(x % 5 == 4){
            if(x > a[5]) a[5] = x;
            flag[5] = false;
        }
    }
    for(int i = 1; i <= 5; i++){
        if(!flag[i]){
            if(i == 4) cout << setprecision(1) << fixed << (a4sum / a[4]);
            else cout << a[i];
        }else{
            cout << "N";
        }
        if(i != 5) cout << " ";
    }
    return 0;
}

1013 考察换行，没有难度

#include<iostream>
#include<math.h>
using namespace std;
bool check(int n){
    bool flag = true;
    for(int i = 2; i <= sqrt(n); i++){
        if(n % i == 0){
            flag = false;
            break;
        }
    }
    return flag;
}
int main(){
    int m, n;
    cin >> m >> n;
    int count = 0;
    int i = 1;
    while(count < n){
        if(check(++i)){
            ++count;
            if(count >= m){
                if((count - m) % 10 != 0) cout << ' ';
                cout << i;
                if((count - m) % 10 == 9)cout <<endl;
            }
            
        }
    }
}

 1016.简单题，考察取位

#include<iostream>
using namespace std;
int main(){
    int a, da, b, db, pa, pb;
    cin >> a >> da >> b >>db;
    pa = 0;
    pb = 0;
    while(a != 0){
        if(a % 10 == da){
            pa = pa * 10 + da;
        }    
        a /= 10;
    }
    while(b != 0){
        if(b % 10 == db){
            pb = pb * 10 + db;
        }    
        b /= 10;
    }
    cout << pa + pb;
    return 0;
}

1017. 简单的除法模拟

#include<iostream>
#include<string>
using namespace std;
int main(){
    string a;
    int b;
    int index, left;
    int first,temp;
    cin >> a >> b;
    first = 0;
    temp = a[0] -'0';
        first = temp / b;
        left = temp % b;
    if(a.size() == 1) cout << first;
    else if(first != 0) cout << first;
    for(int i = 1; i < a.size(); i++){
        temp = left * 10 + (a[i] - '0');
        cout << temp / b;
        left = temp % b;
    }
    cout << ' ' << left;
    return 0;
}

1019. 要用到简单的排序，注意首位有0的情况

#include<iostream>
using namespace std;
int getmax(int a[]){
    for(int i = 1; i < 4; i++){
        for(int j = 0; j < i; j++){
            if(a[j] < a[i]) {
                int n = a[j];
                a[j] = a[i];
                a[i] = n;
            }
        }
    }
    int result = 0;
    for(int i = 0; i < 4; i++){
        result = result * 10 + a[i];
    }
    return result;
}
int getmin(int a[]){
    for(int i = 1; i < 4; i++){
        for(int j = 0; j < i; j++){
            if(a[j] > a[i]) {
                int n = a[j];
                a[j] = a[i];
                a[i] = n;
            }
        }
    }
    int result = 0;
    for(int i = 0; i < 4; i++){
        result = result * 10 + a[i];
    }
    return result;
}
void print(int n){
    int a[4]={0};
    for(int i = 3; i >= 0; i--){
        a[i] = n % 10;
        n /= 10;
    }
    for(int i = 0; i < 4; i++){
        cout << a[i];
    }
}
int main(){
    int n = 0;
    cin >> n;
    int a[4] ={0};
    int result;
    result = -1;
    while(result !=0 && result != 6174){
        for(int i = 0; i < 4; i++){
            a[i] = n % 10;
            n /= 10;        
        }
        int max = getmax(a);
        print(max);
        cout << " - ";
        int min = getmin(a);
        print(min);
        cout << " = ";
        result = max - min;
        print(result);
        cout << endl;
        n = result;
    }

}

1022.简单的取余运算，注意基数为0的情况

#include<iostream>
using namespace std;
void F(long long  n,int m)
{
 int a[10000];
 int index  = 0;
 while(n != 0){
     a[index++] = n % m;
     n = n / m;
 }
 for(int i = index - 1; i >= 0; i--){
     cout << a[i];
 }
 return ;
}
int main(){
    long long n,m;
    int base;
    cin >> n >> m >> base;
    if(n == 0 || m ==0) cout << 0;
    else F(n + m,base);
}