2021-10-09 【PTA】【乙级】【题解5】

1012 数字分类 (20 分)

#include <bits/stdc++.h>
using namespace std;
//1012 数字分类 (20 分)
int main()
{
   
    int A1 = 0, A2 = 0, A3 = 0, A5 = 0;
    double A4 = 0.0;
    int a;
    int k = 1;
    int cntA2 = 0, cntA4 = 0;
    int N;
    cin >> N;

    //for(int i=0;i<N;i++)
    while (N--)
    {
   
        cin >> a;
        if (a % 10 == 0)
            A1 += a;
        else if (a % 5 == 1)
        {
   
            A2 = A2 + k * a;
            k = -k;
            cntA2++;
        }
        else if (a % 5 == 2)
            A3++;
        else if (a % 5 == 3)
        {
   
            A4 += a;
            cntA4 = cntA4 + 1;
        }
        else if (a % 5 == 4)
        {
   
            if (a > A5)
                A5 = a;
        }
    }

    if (cntA4 > 0)
        A4 = A4 / cntA4;

    //输出
    if (A1 == 0)
        cout << "N"
             << " ";
    else if (A1 > 0)
        cout << A1 << " ";

    if (cntA2 == 0)
        cout << "N"
             << " ";
    else if (cntA2 > 0)
        cout << A2 << " ";

    if (A3 == 0)
        cout << "N"
             << " ";
    else if (A3 > 0)
        cout << A3 << " ";

    if (A4 == 0.0)
        cout << "N"
             << " ";
    else if (A4 > 0.0)
        printf("%.1f ", A4);

    if (A5 == 0)
        cout << "N";
    else if (A5 > 0)
        cout << A5;

    return 0;
}

1013 数素数 (20 分)

#include<bits/stdc++.h>
using namespace std;
// 1013 数素数 (20 分)
const int maxn = 200010;
bool judge[maxn]; //判断是否为合数
int pri[maxn], cnt = 0,ans=0;
void getprime(int n)
{
   
    for (int i = 2; i <= n; i++)
    {
   
        if (!judge[i])
            pri[cnt++] = i;
        for (int j = 0; j < cnt; j++)
        {
   
            if (i * pri[j] > n)
                break;
            judge[i * pri[j]] = 1; //标记合数
            if (i % pri[j] == 0)
                break;
        }
    }
}
int main(){
   
    ios::sync_with_stdio(0);
    int n, m;
    cin >> n >> m;
    int cnt = 0;
    getprime(200010);
    for (int i = n; i <= m;i++){
   
        cnt++;
        cout << pri[i - 1];
        if(cnt%10==0||cnt==m-n+1)
            cout << endl;
        else
            cout << " ";
    }
        return 0;
}

1014 福尔摩斯的约会 (20 分)

#include<bits/stdc++.h>
using namespace std;
// 1014 福尔摩斯的约会 (20 分)
int main()
{
   
	int flag = 0;
	string s1, s2, s3, s4;
	cin >> s1 >> s2 >> s3 >> s4;
	string weight[7] = {
    "MON","TUE","WED","THU","FRI","SAT","SUN" };
	for (int i = 0; i < s1.size()&&i<s2.size(); i++)
	{
   
		if (flag)
		{
   
			if (s1[i]<='9'&&s1[i]>='0' && s1[i] == s2[i])
			{
   
				printf("%02d:", (s1[i] - '0'));
				break;
			}
			else if (s1[i] >= 'A'&&s1[i] <= 'N'&&s1[i] == s2[i])
			{
   
				printf("%02d:", (s1[i] - 'A' + 10));
				break;
			}
		}
		else if (s1[i] >= 'A'&&s1[i] <= 'G'&&s1[i] && s1[i] == s2[i])
		{
   
			flag = 1;
			cout << weight[s1[i] - 'A'] << " ";
		}
	}
	for (int i = 0; i < s3.size()&&i<s4.size(); i++)
	{
   
		if(isalpha(s3[i])&&s3[i]==s4[i])//isalpha()判断一个字符是否是字母
		{
   
			printf("%02d",i);
		}
	}
	return 0;
}

1031 查验身份证 (15 分)

#include<bits/stdc++.h>
using namespace std;
//1031 查验身份证 (15 分)
const int weight[17] = {
   7,9,10,5,8,4,2,1,6,3,7,9,10,5,8,4,2};
const char ch[11] = {
   '1','0','X','9','8','7','6','5','4','3','2'};
int check(string s){
   
    int cnt = 0;
    for (int i = 0; i < 17;i++)
        cnt += (s[i] - '0') * weight[i];
    return cnt % 11;
}
int main(){
   
    ios::sync_with_stdio(0);
    int n, ans = 0, cnt = 0;
    string id;
    cin >> n;
    for (int i = 0; i < n;i++){
   
        cin >> id;
        ans = check(id);
        if(ch[ans]==id[17])
            cnt++;
        else
            cout << id << endl;
    }
    if(cnt==n)
        cout << "All passed";
        return 0;
}

1017 A除以B (20 分)

#include<bits/stdc++.h>
using namespace std;
//1017 A除以B (20 分)
int main(){
   
    ios::sync_with_stdio(0);
    string s;
    int n, t = 0, temp = 0;
    cin >> s >> n;
    int len = s.size();
    t = (s[0] - '0') / n;
    if((t!=0&&len>1)||len==1)
        cout << t;
    temp = (s[0] - '0') % n;
    for (int i = 1; i < len;i++){
   
        t = (temp * 10 + s[i] - '0') / n;
        cout << t;
        temp = (temp * 10 + s[i] - '0') % n;
    }
    cout << " " << temp;
    return 0;
}