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Codeforces Round 957 (Div.3)

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7月11日 Codeforces 实战记录。（1461 -> 1505）

AC数5/7。
比赛跳转：Codeforces Round 957 (Div. 3)

题解

A. Only Pluses

分析：每次给最小的数 +1。

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

inline int read(){
    int x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') { if(ch == '-') f = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

void solve(){
    vector<int>v(3);
    cin >> v[0] >> v[1] >> v[2];
    for(int i = 1; i <= 5; i++){
        sort(v.begin(), v.end());
        v[0]++;
    }
    cout << v[0] * v[1] * v[2] << '\n';
}

signed main(){
    int t = read();
    while(t--) solve();
    return 0;
}

B. Angry Monk

分析：保留最大数，其他数加入最大数，n - 1 次拆成全1，n 次合并，对每个数计 2n - 1 次。

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

inline int read(){
    int x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') { if(ch == '-') f = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

void solve(){
    int n = read(), k = read();
    vector<int>v(k);
    for(auto &i : v) i = read();
    
    sort(v.begin(), v.end());
    int res = 0;
    for(int i = 0; i < k - 1; i++){
        res += v[i] + v[i] - 1;
    }
    cout << res << '\n';
}

signed main(){
    int t = read();
    while(t--) solve();
    return 0;
}

C. Gorilla and Permutation

分析：构造题，给 >= k 的数分配尽可能大的权重，放在序列开头部分（降序）， <= m 的数分配尽可能小的权重，放在序列尾端部分（升序）

如果一个数同时满足这两个条件，则不会被计入权重，应该将其放在序列中间部分，保证尾端部分的权重更小。

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

inline int read(){
    int x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') { if(ch == '-') f = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

void solve(){
    int n = read(), m = read(), k = read();
    int mid = min(k - 1, m);
    for(int i = n; i > mid; i--){
        cout << i << ' ';
    }
    for(int i = 1; i <= mid; i++){
        cout << i << ' ';
    }
    puts("");
}

signed main(){
    int t = read();
    while(t--) solve();
    return 0;
}

D. Test of Love

模拟题：贪心，先尽可能跳到 L 上，如果不行，跳到尽可能远的 W 里，游动，重复此过程。

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

inline int read() {
    int x = 0, f = 1; char ch = getchar();
    while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

void solve() {
    int n = read(), m = read(), k = read();
    string s; cin >> s; s = "L" + s;

    int cntw = 0;
    for (int i = 0; i <= n;) {
        
        int ok = 0;
        for (int j = m; j > 0; j--) {
            if (s[i + j] == 'L') {
                i += j;
                ok = 1;
                break;
            }
        }
        
        if (i + m >= n + 1) {
            cout << "YES" << '\n';
            return;
        }

        if (!ok) {
            for (int j = m; j > 0; j--) {
                if (s[i + j] == 'W') {
                    i += j;
                    ok = 2;
                    break;
                }
            }
        }

        if (ok == 0) {
            cout << "NO" << '\n';
            return;
        }
        else if (ok == 1) {
            continue;
        }

        while (s[i] == 'W') {
            i++;
            cntw++;
            if (cntw > k || s[i] == 'C') {
                cout << "NO" << '\n';
                return;
            }
            if (i == n + 1) {
                cout << "YES" << '\n';
                return;
            }
        }
    }
}

signed main() {
    int t = read();
    while (t--) solve();
    return 0;
}

E. Novice’s Mistake

分析：模拟题意即可，注意控制 b 的范围避免不必要情况。

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

inline int read(){
    int x = 0, f = 1; char ch = getchar();
    while(ch < '0' || ch > '9') { if(ch == '-') f = -1; ch = getchar(); }
    while(ch >= '0' && ch <= '9') { x = x * 10 + ch - 48; ch = getchar(); }
    return x * f;
}

int get(string& str, int len) {
    int x = 0;
    for (int i = 0; i < len; i++) {
        x = x * 10 + str[i] - 48;
    }
    return x;
}

void solve(){
    int n = read();
    int cnt = 0;
    vector<pair<int, int>> ans;
    string s = to_string(n);
    string str = "";
    for (int a = 1; a <= 10000; a++) {
        str += s;

        int len = str.length();
        int maxb = min({ (long long)10000, a * n, len - 1 });
        for (int b = max((long long)1, len - 10); b <= maxb; b++) {
            if (get(str, len - b) == a * n - b) {
                cnt++;
                ans.emplace_back(a, b);
            }
        }
    }
    cout << cnt << '\n';
    for(auto [a, b] : ans){
        cout << a << ' ' << b << '\n';
    }
}

signed main(){
    int t = read();
    while(t--) solve();
    return 0;
}