【PTA】乙级题解 11 -20

最新推荐文章于 2022-11-13 13:34:29 发布

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1011 A+B 和 C

#include <iostream>
using namespace std;
int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
        long long int a, b, c;
        scanf("%lld%lld%lld", &a, &b, &c);
        printf("Case #%d: %s\n", i + 1, a + b > c ? "true" : "false");
    }
    return 0;
}

1012 数字分类


#include <iostream>
#include <vector>
using namespace std;
int main() {
    int n, num, A1 = 0, A2 = 0, A5 = 0;
    double A4 = 0.0;
    cin >> n;
    vector<int> v[5];
    for (int i = 0; i < n; i++) {
        cin >> num;
        v[num%5].push_back(num);
    }
    for (int i = 0; i < 5; i++) {
        for (int j = 0; j < v[i].size(); j++) {
            if (i == 0 && v[i][j] % 2 == 0) A1 += v[i][j];
            if (i == 1 && j % 2 == 0) A2 += v[i][j];
            if (i == 1 && j % 2 == 1) A2 -= v[i][j];
            if (i == 3) A4 += v[i][j];
            if (i == 4 && v[i][j] > A5) A5 = v[i][j];
        }
    }
    for (int i = 0; i < 5; i++) {
        if (i != 0) printf(" ");
        if (i == 0 && A1 == 0 || i != 0 && v[i].size() == 0) {
            printf("N"); continue;
        }
        if (i == 0) printf("%d", A1);
        if (i == 1) printf("%d", A2);
        if (i == 2) printf("%d", v[2].size());
        if (i == 3) printf("%.1f", A4 / v[3].size());
        if (i == 4) printf("%d", A5);
    }
    return 0;
}

1013 数素数 (20分)

#include <iostream>
#include <vector>
using namespace std;
bool isprime(int a) {
    for (int i = 2; i * i <= a; i++)
        if(a % i == 0) return false;
    return true;
}
int main() {
    int M, N, num = 2, cnt = 0;
    cin >> M >> N;
    vector<int> v;
    while (cnt < N) {
        if (isprime(num)) {
            cnt++;
            if (cnt >= M) v.push_back(num);
        }
        num++;
    }
    cnt = 0;
    for (int i = 0; i < v.size(); i++) {
        cnt++;
        if (cnt % 10 != 1) printf(" ");
        printf("%d", v[i]);
        if (cnt % 10 == 0) printf("\n");
    }
    return 0;
}

1014. 福尔摩斯的约会
分析：按照题目所给的方法找到相等的字符后判断即可，如果输出的时间不足2位数要在前面添0，即用%02d输出～

#include <iostream>
#include <cctype>
using namespace std;
int main() {
    string a, b, c, d;
    cin >> a >> b >> c >> d;
    char t[2];
    int pos, i = 0, j = 0;
    while(i < a.length() && i < b.length()) {
        if (a[i] == b[i] && (a[i] >= 'A' && a[i] <= 'G')) {
            t[0] = a[i];
            break;
        }
        i++;
    }
    i = i + 1;
    while (i < a.length() && i < b.length()) {
        if (a[i] == b[i] && ((a[i] >= 'A' && a[i] <= 'N') || isdigit(a[i]))) {
            t[1] = a[i];
            break;
        }
        i++;
    }
    while (j < c.length() && j < d.length()) {
        if (c[j] == d[j] && isalpha(c[j])) {
            pos = j;
            break;
        }
        j++;
    }
    string week[7] = {"MON ", "TUE ", "WED ", "THU ", "FRI ", "SAT ", "SUN "};
    int m = isdigit(t[1]) ? t[1] - '0' : t[1] - 'A' + 10;
    cout << week[t[0]-'A'];
    printf("%02d:%02d", m, pos);
    return 0;
}

1015. 德才论
分析：用结构体存储。写好cmp函数~结构体数组vector v[4]中v[0]保存第一类考生，v[1]保存第二类考生…以此类推。写好cmp函数很重要，cmp函数中，排序先按照总分排序，然后按照德分排序，最后按照才分排序…最后输出符合条件的结果～

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
struct node 
{
    int num, de, cai;
};
int cmp(struct node a, struct node b) 
{
    if ((a.de + a.cai) != (b.de + b.cai))
        return (a.de + a.cai) > (b.de + b.cai);
    else if (a.de != b.de)
        return a.de > b.de;
    else
        return a.num < b.num;
}
int main() 
{
    int n, low, high;
    scanf("%d %d %d", &n, &low, &high);
    vector<node> v[4];
    node temp;
    int total = n;
    for (int i = 0; i < n; i++)
     {
        scanf("%d %d %d", &temp.num, &temp.de, &temp.cai);
        if (temp.de < low || temp.cai < low)
            total--;
        else if (temp.de >= high && temp.cai >= high)
            v[0].push_back(temp);
        else if (temp.de >= high && temp.cai < high)
            v[1].push_back(temp);
        else if (temp.de < high && temp.cai < high && temp.de >= temp.cai)
            v[2].push_back(temp);
        else
            v[3].push_back(temp);
    }
    printf("%d\n", total);
    for (int i = 0; i < 4; i++)
     {
        sort(v[i].begin(), v[i].end(), cmp);
        for (int j = 0; j < v[i].size(); j++)
            printf("%d %d %d\n", v[i][j].num, v[i][j].de, v[i][j].cai);
    }
    return 0;
}

1016. 部分A+B


#include <iostream>
using namespace std;
int main() {
    string a, b;
    int da, db, pa = 0, pb = 0;
    cin >> a >> da >> b >> db;
    for (int i = 0; i < a.length(); i++)
        if (da == (a[i] - '0')) pa = pa * 10 + da;
    for (int i = 0; i < b.length(); i++)
        if (db == (b[i] - '0')) pb = pb * 10 + db;
    cout << pa + pb;
    return 0;
}

1017. A除以B

#include <iostream>
using namespace std;
int main() {
    string s;
    int a, t = 0, temp = 0;
    cin >> s >> a;
    int len = s.length();
    t = (s[0] - '0') / a;
    if ((t != 0 && len > 1) || len == 1)
        cout << t;
    temp = (s[0] - '0') % a;
    for (int i = 1; i < len; i++) {
        t = (temp * 10 + s[i] - '0') / a;
        cout << t;
        temp = (temp * 10 + s[i] - '0') % a;
    }
    cout << " " << temp;
    return 0;
}

1018. 锤子剪刀布


#include <iostream>
using namespace std;
int main() {
    int n;
    cin >> n;
    int jiawin = 0, yiwin = 0;
    int jia[3] = {0}, yi[3] = {0};
    for (int i = 0; i < n; i++) {
        char s, t; 这个char变量的巧妙！
        cin >> s >> t;
        if (s == 'B' && t == 'C') {
            jiawin++;
            jia[0]++;
        } else if (s == 'B' && t == 'J') {
            yiwin++;
            yi[2]++;
        } else if (s == 'C' && t == 'B') {
            yiwin++;
            yi[0]++;
        } else if (s == 'C' && t == 'J') {
            jiawin++;
            jia[1]++;
        } else if (s == 'J' && t == 'B') {
            jiawin++;
            jia[2]++;
        } else if (s == 'J' && t == 'C') {
            yiwin++;
            yi[1]++;
        }
    }
    cout << jiawin << " " << n - jiawin - yiwin << " " << yiwin << endl << yiwin << " " << n - jiawin - yiwin << " " << jiawin << endl;
    int maxjia = jia[0] >= jia[1] ? 0 : 1; // 这里是大于等于号，注意！
    maxjia = jia[maxjia] >= jia[2] ? maxjia : 2;
    int maxyi = yi[0] >= yi[1] ? 0 : 1;
    maxyi = yi[maxyi] >= yi[2] ? maxyi : 2;
    char str[4] = {"BCJ"}; //通过这个实现字母序排序
    cout << str[maxjia] << " " << str[maxyi];
    return 0;
}

1019 数字黑洞

#include <iostream>
#include <algorithm>
using namespace std;
bool cmp(char a, char b) {return a > b;}
int main() {
    string s;
    cin >> s;
    s.insert(0, 4 - s.length(), '0');
    do {
        string a = s, b = s;
        sort(a.begin(), a.end(), cmp);
        sort(b.begin(), b.end());
        int result = stoi(a) - stoi(b);
        s = to_string(result);
        s.insert(0, 4 - s.length(), '0');
        cout << a << " - " << b << " = " << s << endl;
    } while (s != "6174" && s != "0000");
    return 0;
}

1020 月饼


#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
struct mooncake{
    float mount, price, unit;
};
int cmp(mooncake a, mooncake b) {
    return a.unit > b.unit;
}
int main() {
    int n, need;
    cin >> n >> need;
    vector<mooncake> a(n); 思考这个地方为什么要用vector数组
    用vector数组解决了多个结构体的输入问题！
    for (int i = 0; i < n; i++) scanf("%f", &a[i].mount);
    for (int i = 0; i < n; i++) scanf("%f", &a[i].price);
    for (int i = 0; i < n; i++) a[i].unit = a[i].price / a[i].mount;
    sort(a.begin(), a.end(), cmp);
    float result = 0.0;
    for (int i = 0; i < n; i++) {
        if (a[i].mount <= need) {
            result = result + a[i].price;
        } else {
            result = result + a[i].unit * need;
            break;
        }
        need = need - a[i].mount;
    }
    printf("%.2f",result);
    return 0;
}

补充知识

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main ()
{
    char str[]="1776ad";
    int year;
    if (isdigit(str[0]))
    {
        year = atoi (str);
        printf ("The year that followed %d was %d.\n", year, year+1);
    }
    return 0;
}