Day1

板刷CF构造题

A. Nastia and Nearly Good Numbers

https://codeforces.com/problemset/problem/1521/A
构造\(z=abt_3,y=at_2,x=at_1\)，根据等式\(z=x+y\)，可以约去a
故可以构造 1 3b-1 3b （系数）
注意特判b==1，全部都是good，无法构造

#include<bits/stdc++.h>

using namespace std;

void solve () {
    long long a, b;
    cin >> a >> b;
    if (b == 1) {
        cout << "NO\n";
        return ;
    }
    cout << "YES\n";
    cout << a << ' ' << a * (3*b-1) << ' ' << a*3*b << endl;
    
    
}

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

//x+y=z
//一个%ab==0,另两个%a==0
//t1+t2==bt3
//1 b-1 b

A. K-divisible Sum

https://codeforces.com/problemset/problem/1476/A
懂抽屉原理的宝宝(￣︶￣*))就可以秒解啦
小心格式的输出

#include <bits/stdc++.h>

using namespace std;

void solve () {
    int n, k;
    cin >> n >> k;
    if (n == 1)         cout << k << endl;
    else if (n == k)    cout << 1 << endl;
    else if (n < k)     cout << (int)ceil (1.0*k/n) << endl;
    else {
        double dx = n - k;
        //cout << dx << ' ';
        k = ceil(dx/k)*k+k;
        //cout << k << ' ';
        cout << (int)ceil (1.0*k/n) << endl;
    }
}

int main () {
    int t;
    cin >> t;
    while (t --) {
        solve();
    }
}
//抽屉原理
//啊啊啊气死我了！！输出要注意格式int

C. Not Adjacent Matrix

https://codeforces.com/problemset/problem/1520/C
灵光乍现，根据对角线来填充，这样子就不会有相邻的在上下左右了。
注意怎么构造：找规律+普适性规律

#include <bits/stdc++.h>

using namespace std;
const int N = 105;
int a[N][N];

void print (int n) {
    for (int i = 1; i <= n; i ++) {
        for (int j = 1; j <= n; j ++)
            cout << a[i][j] << ' ';
        cout << endl;
    }
}

void solve () {
    int n;
    cin >> n;
    if (n == 1)     cout << "1\n";
    else if (n == 2)    cout << "-1\n";
    else {
        a[1][1] = 1;
        //处理第一行, 增量为 1*n, 2*(n-i+2)...
        a[1][2] = a[1][1] + n;
        for (int i = 3; i <= n; i ++)   a[1][i] = a[1][i-1] + 2*(n-i+2);
        //处理第一列 增量为 ...9,7,5,3 通项:2*(n-i)+3
        for (int i = 2; i <= n; i ++)   a[i][1] = a[i-1][1] + 2*(n-i)+3;
        //然后就可以根据对角线求啦
        for (int i = 2; i <= n; i ++)
            for (int j = 2; j <= n; j ++)
                a[i][j] = a[i-1][j-1] + 1;

        print (n);
    }
}

int main () {
    int t;
    cin >> t;
    while (t --) {
        solve ();
    }
}
//上下左右不得相邻
//按照对角线填
//从主对角线开始填，然后一上一下挨个填

B. Same Parity Summands

https://codeforces.com/problemset/problem/1352/B
奇偶性相同的k个数和为n
构造:
1 1 1 1...n-k+1
2 2 2 2...n-2*(k-1)
//构造没有问题，但是要记得判断末项是否大于0

#include <bits/stdc++.h>

using namespace std;

void solve () {
    int n, k;
    cin >> n >> k;
    if (n % k == 0) {
        cout << "YES" << endl;
        for (int i = 0; i < k; i ++)
            cout << n / k << ' ';
        cout << endl;
    }
    else {
        int res1 = n-k+1, res2 = n-2*(k-1);
        if (res1 > 0 && res1 & 1) {
            cout << "YES\n";
            for (int i = 0; i < k - 1; i ++)   cout << "1 ";
            cout << n-k+1 << endl;
            return ;
        }
        if (res2 > 0 && res2 % 2 == 0) {
            cout << "YES\n";
            for (int i = 0; i < k - 1; i ++)    cout << "2 ";
            cout << n - 2 * (k-1) << endl;
            return ;
        }
        cout << "NO\n";
    }

}

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

//奇偶性相同的k个数和为n
//构造:
// 1 1 1 1...n-k+1
// 2 2 2 2...n-2*(k-1)
//构造没有问题，但是要记得判断末项是否大于0

标签：...,cout,int,菜鸟,构造,Day1,ACMer,solve,include
来源： https://www.cnblogs.com/CTing/p/16437455.html