PAT甲级打卡-1005-1010

1005 Spell It Right

给一个大数要求出每一位的和并以英文输出

#include <bits/stdc++.h>

using namespace std;

map<int, string> mp;

signed main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

    mp[0] = "zero";
    mp[1] = "one";
    mp[2] = "two";
    mp[3] = "three";
    mp[4] = "four";
    mp[5] = "five";
    mp[6] = "six";
    mp[7] = "seven";
    mp[8] = "eight";
    mp[9] = "nine";

    string s;
    cin >> s;
    int sum = 0;
    for (auto x : s) {
        sum += (int) (x - '0');
    }
    vector<int> nums;
    if (sum == 0) nums.push_back(0);
    while (sum) {
        nums.push_back(sum % 10);
        sum /= 10;
    }
    reverse(nums.begin(), nums.end());

    for (int i = 0; i < nums.size(); ++i) {
        int x = nums[i];
        cout << mp[x] << (i == nums.size() - 1 ? "" : " ");
    }
}

1006 Sign In and Sign Out

n个人打卡上下班，问最早到的人和最晚下班的人

#include <bits/stdc++.h>

using namespace std;

const int N = 100010;

struct node {
    string name;
    int start;
    int end;
};

node persons[N];

int getTime(string s) {
    int hh = (s[0] - '0') * 10 + (s[1]- '0');
    int mm = (s[3] - '0') * 10 + (s[4]- '0');
    int ss = (s[6] - '0') * 10 + (s[7]- '0');
    return hh * 3600 + mm * 60 + ss;
}

signed main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

    int n;
    cin >> n;
    for(int i = 0; i < n; ++ i) {
        string name, tmps, tmpe;
        cin >> name >> tmps >> tmpe;
        persons[i].name = name;
        persons[i].start = getTime(tmps);
        persons[i].end = getTime(tmpe);
    }

    sort(persons, persons +n,[&](auto x, auto y){
        return x.start < y.start;
    });

    cout << persons[0].name << " ";

    sort(persons, persons +n,[&](auto x, auto y){
        return x.end > y.end;
    });

    cout << persons[0].name;
}

1007 Maximum Subsequence Sum

给n个数问最大连续子区间和，但注意特判细节

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const ll N = 10010;

ll a[N];


signed main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

    ll n;
    cin >> n;
    for (ll i = 1; i <= n; ++i) {
        cin >> a[i];
    }

    ll pos = 1, len = 0;
    ll sum = 0;

    ll res = 0, resl = 1, resr = 1;
    for (ll i = 1; i <= n; ++i) {
        sum += a[i];
        len++;
        if (sum <= 0) {
            pos = i + 1;
            len = 0;
            sum = 0;
        } else {
            if (sum > res) {
                res = sum;
                resl = pos;
                resr = pos + len - 1;
            }
        }
    }

    bool flag = 1;
    for(int i = 1; i<= n; ++ i) {
        if(a[i]>= 0) flag = 0;
    }
    if(flag)  cout << 0 << " " << a[1] << " " << a[n] << endl;
    else if(res == 0) cout << 0 << " "<< 0 << " "<< 0 << endl;
    else cout << res << " " << a[resl] << " " << a[resr] << endl;
    return 0;
}

1008 Elevator

按顺序坐电梯，小模拟

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const ll N = 10010;

signed main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

    ll n;
    cin >> n;
    int p = 0;
    int ans = 5 * n;
    for(int i = 1; i<= n; ++ i) {
        int x;
        cin >> x;
        if(x < p) ans += abs(x - p) * 4;
        else ans += abs(x - p) * 6;
        p = x;
    }
    cout << ans << endl;
    return 0;
}

1009 Product of Polynomials

求两个多项式的乘积，按格式输出

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const ll N = 10010;

struct node {
    int exp;
    double num;
};

vector<node> a, b;

map<int, double> mp;

signed main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);

    ll n;
    cin >> n;
    for (int i = 0; i < n; ++i) {
        int x;
        double k;
        cin >> x >> k;
        a.push_back({x, k});
    }
    cin >> n;
    for (int i = 0; i < n; ++i) {
        int x;
        double k;
        cin >> x >> k;
        b.push_back({x, k});
    }
    for (auto[ax, ak] : a) {
        for (auto[bx, bk] : b) {
            mp[ax + bx] += ak * bk;
        }
    }

    vector<node> res;
    for (auto[x, k] : mp) {
        if (fabs(k) < 1e-7) continue;
        res.push_back({x, k});
    }
    reverse(res.begin(), res.end());

    cout << res.size();
    for (auto[x, k] : res) {
        cout << " " << x << " " << fixed << setprecision(1) << k;
    }
    return 0;
}

1010 Radix

给两个数，给出其中一个数的进制，问这两个数是否可能相等，输出另一个数的进制。

进制问题有一个注意点就是出现的数字要比进制小

对不起我刚开始觉得进制只会有二三十，wa了发

实际上进制可能会非常大，是给出数十进制上限这种级别 1e15，只能二分

不过不二分也有24分）

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
const ll N = 10010;

ll getNum(char ch) {
    if (ch >= '0' && ch <= '9') return (ll) (ch - '0');
    return (ll) (ch - 'a' + 10);
}

__int128 getDecimal(string x, ll p) {
    __int128 res = 0;
    ll pp = 1;
    while (x.size()) {
        ll num = getNum(x.back());
        x.pop_back();
        res += num * pp;
        pp = pp * p;
    }
    return res;
}

signed main() {
    ios::sync_with_stdio(false), cin.tie(nullptr), cout.tie(nullptr);


    string x, y;
    ll op, p;
    cin >> x >> y >> op >> p;
    if (op == 2) swap(x, y);

    __int128 nummm = getDecimal(x, p);

    ll mx = 0;
    for (auto ch : x) {
        mx = max(mx, getNum(ch));
    }

    if (mx >= p) {
        cout << "Impossible";
        return 0;
    }

    mx = 0;
    for (auto ch : y) {
        mx = max(mx, getNum(ch));
    }

//    bool flag = 0;
//    for (ll i = max(2ll, mx + 1); i <= 1000000000; ++i) {
//        __int128 tmp = getDecimal(y, i);
//        if (nummm == tmp) {
//            cout << i;
//            flag = 1;
//            break;
//        }
//        if (tmp > nummm || tmp < 0) break;
//    }

    ll l = mx + 1, r = 1e16, ans = -1;
    while(l <= r) {
        ll mid = (l + r) >> 1;
        __int128 tmp = getDecimal(y, mid);
        if(nummm <= tmp || tmp < 0) {
            r = mid - 1;
            if(nummm == tmp )ans = mid;
        }
        else {
            l = mid + 1;
        }
    }

    if (ans == -1) cout << "Impossible" << endl;
    else cout << ans << endl;

    return 0;
}