CodeForces 1521 E. Nastia and a Beautiful Matrix

首先想到的是对角线构造，接着就构造出了这种情况：

就是考虑让数量多的先填\(1\)，后面接着填。但是这样有会个问题：

这样并不满足条件。对于这个问题，一开始我认为是要再加大面积，后来发现这样与下发的样例的\(p\)不匹配。然后试着对上图调整一下，确实能填下，看来是填的方法出了问题。

有构造了一会发现了以下图形：

x表示可以填任何数，这个成立的条件是"\(1\not=2\)"。

接着考虑填放顺序，首先数量最多的填\(1\)，接着为了使\(1\not=2\)继续填x，最后填\(2\)，通过反证法可知一定合法。

#include <bits/stdc++.h>

namespace io {
    const int SIZE = (1 << 21) + 1;
    char ibuf[SIZE], *iS, *iT, obuf[SIZE], *oS = obuf, *oT = oS + SIZE - 1, c, qu[55]; int f, qr;
    // getchar
    #define gc() (iS == iT ? (iT = (iS = ibuf) + fread (ibuf, 1, SIZE, stdin), (iS == iT ? EOF : *iS ++)) : *iS ++)
    // print the remaining part
    inline void flush () {
        fwrite (obuf, 1, oS - obuf, stdout);
        oS = obuf;
    }
    // putchar
    inline void putc (char x) {
        *oS ++ = x;
        if (oS == oT) flush ();
    }
    // input a signed integer
    template <class I>
    inline void gi (I &x) {
        for (f = 1, c = gc(); c < '0' || c > '9'; c = gc()) if (c == '-') f = -1;
        for (x = 0; c <= '9' && c >= '0'; c = gc()) x = x * 10 + (c & 15); x *= f;
    }
    // print a signed integer
    template <class I>
    inline void print (I x) {
        if (!x) putc ('0'); if (x < 0) putc ('-'), x = -x;
        while (x) qu[++ qr] = x % 10 + '0',  x /= 10;
        while (qr) putc (qu[qr --]);
    }
    //no need to call flush at the end manually!
    struct Flusher_ {~Flusher_(){flush();}}io_flusher_;
}
using io :: gi;
using io :: putc;
using io :: print;

int main(void) {
	freopen("aztec.in", "r", stdin), freopen("aztec.out", "w", stdout);
	
	int testCase = 1;
	gi(testCase);
	while (testCase--) {
		int N, M, mx = 0;
		gi(N), gi(M);
		std::vector <int> A(N + 1), E(N + 1, 0);
		for (int i = 1; i <= N; ++i) {
			gi(A[i]), E[i] = i, mx = std::max(mx, A[i]);
		}
		
		std::sort(E.begin() + 1, E.end(), [&](const int& a, const int& b) { return A[a] > A[b]; });
		
		auto ok = [&](int p) {
			if ((p + 1 >> 1) * p < mx)
				return false;
			if (p * p - (p >> 1) * (p >> 1) < M)
				return false;
			return true;
		};
		
		int p = 1;
		while (!ok(p))
			++p;
		print(p), putc('\n');
		std::vector <std::vector <int>> R(p + 1, std::vector <int> (p + 1, 0));
		
		if (N == 1) {
			for (int i = 1; i <= p; i += 2) {
				for (int j = 1; j <= p; ++j) {
					if (A[1])
						R[i][j] = 1, --A[1];
				}
			}
			
			for (int i = 1; i <= p; ++i)  {
				for (int j = 1; j <= p;++j)
					print(R[i][j]), putc(" \n"[j == p]);
			}
			continue;
		}
		
		int o = 1;
		for (int i = 1; i <= p; i += 2) {
			for (int j = p & 1 ? 2 : 1; j <= p; j += 2) {
				if (!A[E[o]])
					++o;
				if (o > N || !A[E[o]])
					break;
				--A[E[o]], R[i][j] = E[o];
			}
		}
		
		for (int i = 1; i <= p; i += 2) {
			for (int j = p & 1 ? 1 : 2; j <= p; j += 2) {
				if (o > N)
					break;
				if (!A[E[o]])
					++o;
				if (o > N || !A[E[o]])
					break;
				--A[E[o]];
				R[i][j] = E[o];				
			}
		}
		
		for (int i = 2; i <= p; i += 2) {
			for (int j = p & 1 ? 1 : 2; j <= p; j += 2) {
				if (o > N)
					break;
				if (!A[E[o]])
					++o;
				if (o > N || !A[E[o]])
					break;
				--A[E[o]];
				R[i][j] = E[o];
			}
		}
		
		for (int i = 1; i <= p; ++i) {
			for (int j = 1; j <= p; ++j) {
				print(R[i][j]), putc(" \n"[j == p]);
			}
		}
	}
	
	return 0;
}