Codeforces Round #640 (Div. 4)

Posted 2020-11-28 kanoon

比赛链接：https://codeforces.com/contest/1352

A - Sum of Round Numbers

题意

将一个十进制数的每一个非零位分离出来。

代码

#include <bits/stdc++.h>
using namespace std;

void solve() {
    string s; cin >> s;
    int n = s.size();
    vector<string> ans;
    for (int i = n - 1; i >= 0; i--) {
        if (s[i] != ‘0‘) {
            ans.push_back(s[i] + string(n - 1 - i, ‘0‘));
        }
    }
    cout << ans.size() << "
";
    for (auto i : ans) cout << i << ‘ ‘;
    cout << "
";
}

int main() {
    int t; cin >> t;
    while (t--) solve();
}

B - Same Parity Summands

题意

将整数 n 分解为 k 个奇偶性相同的数。

代码

#include <bits/stdc++.h>
using namespace std;

void solve() {
    int n, k; cin >> n >> k;
    int n1 = n - (k - 1), n2 = n - (k - 1) * 2;
    if (n1 > 0 and n1 % 2 == 1) {
        cout << "YES
";
        for (int i = 0; i < k - 1; i++) cout << "1 ";
        cout << n1 << "
";
    } else if (n2 > 0 and n2 % 2 == 0) {
        cout << "YES
";
        for (int i = 0; i < k - 1; i++) cout << "2 ";
        cout << n2 << "
";
    } else {
        cout << "NO" << "
";
    }
}

int main() {
    int t; cin >> t;
    while (t--) solve();
}

C - K-th Not Divisible by n

题意

输出第 k 个不能被 n 整除的数。

题解

两个 n 之间有 n - 1 个不能被 n 整除的数，找出 k 中有多少个完整的 n - 1 区间，再对 n - 1 取余即可。

代码

#include <bits/stdc++.h>
using namespace std;

void solve() {
    long long n, k; cin >> n >> k;
    if (k % (n - 1) == 0)
        cout << k / (n - 1) * n - 1 << "
";
    else
        cout << k / (n - 1) * n + k % (n - 1) << "
";
}

int main() {
    int t; cin >> t;
    while (t--) solve();
}

D - Alice, Bob and Candies

题意

n 堆糖果排成一排，Alice 只能从左边取，Bob 只能从右边取，每次一人取得的糖果数需多于另一人上一次取得的糖果数，问所有糖果取完时二人共取了多少次，以及各自取得的糖果个数。

题解

模拟即可。

代码

#include <bits/stdc++.h>
using namespace std;

void solve() {
    int n; cin >> n;
    int a[n] = {}; for (auto &i : a) cin >> i;
    int l = 0, r = n - 1;
    int ans_l = 0, ans_r = 0;
    int pre_l = 0, pre_r = 0;
    int i;
    for (i = 0; l <= r; i++) {
        int sum = 0;
        if (i % 2 == 0) {
            while (l <= r and sum <= pre_r) {
                sum += a[l];
                ++l;
            }
            ans_l += sum;
            pre_l = sum;
        } else {
            while (r >= l and sum <= pre_l) {
                sum += a[r];
                --r;
            }
            ans_r += sum;
            pre_r = sum;
        }
    }
    cout << i << ‘ ‘ << ans_l << ‘ ‘ << ans_r << "
";
}

int main() {
    int t; cin >> t;
    while (t--) solve();
}

E - Special Elements

题意

统计数组 a 中有多少元素可以被连续子数组加和求得。

题解

数据范围较小，$O_{(n^2)}$ 枚举即可。

代码

#include <bits/stdc++.h>
using namespace std;

const int M = 8100;
int n, a[M], cnt[M];

void solve() {
    fill(cnt, cnt + M, 0);
    cin >> n;
    for (int i = 0; i < n; i++) {
        cin >> a[i], ++cnt[a[i]];
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
        int sum = a[i];
        for (int j = i + 1; j < n; j++) {
            sum += a[j];
            if (sum < M and cnt[sum]) {
                ans += cnt[sum];
                cnt[sum] = 0;
            }
        }
    }
    cout << ans << "
";
}

int main() {
    int t; cin >> t;
    while (t--) solve();
}

F - Binary String Reconstruction

题意

构造一个 01 串，要求：

• 00 的个数为 n0 个
• 01 的个数为 n1 个
• 11 的个数为 n2 个

题解

依次构造 11，10，00 即可。

代码

#include <bits/stdc++.h>
using namespace std;

void solve() {
    int n0, n1, n2; cin >> n0 >> n1 >> n2;
    vector<int> ans;
    if (n1 or n2) ans.push_back(1);
    for (int i = 0; i < n2; i++) ans.push_back(1);
    for (int i = 0; i < n1 / 2; i++) {
        ans.push_back(0);
        ans.push_back(1);
    }
    if (n1 & 1) {
        ans.push_back(0);
        for (int i = 0; i < n0; i++) ans.push_back(0);
    } else {
        if (ans.size()) {
            ans.pop_back();
            for (int i = 0; i < n0; i++) ans.push_back(0);
            ans.push_back(1);
        } else {
            for (int i = 0; i < n0 + 1; i++) ans.push_back(0);
        }
    }
    for (auto i : ans) cout << i;
    cout << "
";
}

int main() {
    int t; cin >> t;
    while (t--) solve();
}

G - Special Permutation

题意

构造数 n 的一个排列，使得相邻两数相差在 [2, 4] 之间。

题解

顺着 3 1 4 2 构造即可。

代码

#include <bits/stdc++.h>
using namespace std;

void solve() {
    int n; cin >> n;
    if (n <= 3) {
        cout << -1 << "
";
        return;
    } else {
        for (int i = n - (n % 2 == 0); i >= 5; i -= 2) cout << i << ‘ ‘;
        cout << "3 1 4 2 ";
        for (int i = 6; i <= n - (n % 2 == 1); i += 2) cout << i << ‘ ‘;
        cout << "
";
    }
}

int main() {
    int t; cin >> t;
    while (t--) solve();
}