【CodePlus 2017 12 月赛】可做题1 & LibreOJ6256

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Description

click me

Solution

可以证明每个矩阵是巧妙的当且仅当其每个2阶子矩阵均是巧妙的：
若该矩阵有一个不巧妙的2阶子矩阵，则其他部分任意选择，这两行选择不同，则和不同，所以该矩阵不是巧妙的。
用goodi,j表示以(i,j)为左上角的二阶子矩阵是否巧妙，计算前缀和即可。

Code

/**************************Au: Hany01Date: Dec 24th, 2017Prob: Code+ Round2 Div2 T2Email: hany01@foxmail.com**************************/#include<bits/stdc++.h>using namespace std;typedef long long ll;typedef pair<int, int> PII;#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)#define Set(a, b) memset(a, b, sizeof(a))#define Cpy(a, b) memcpy(a, b, sizeof(a))#define fir first#define sec second#define pb(a) push_back(a)#define mp(a, b) make_pair(a, b)#define ALL(a) (a).begin(), (a).end()#define SZ(a) ((int)(a).size())#define INF (0x3f3f3f3f)#define INF1 (2139062143)#define Ha (1000000007)template <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }inline int read(){    int _, __; char c_;    for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;    for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);    return _ * __;}inline void File(){#ifdef hany01    freopen("solve1.in", "r", stdin);    freopen("solve1.out", "w", stdout);#endif}const int maxn = 505;int n, m, T, a[maxn][maxn], good[maxn][maxn], x, y, k;int main(){    File();    n = read(); m = read(); T = read();    For(i, 1, n) For(j, 1, m) a[i][j] = read();    For(i, 1, n - 1) For(j, 1, m - 1)        good[i][j] = good[i - 1][j] + good[i][j - 1] - good[i - 1][j - 1] + (a[i][j] + a[i + 1][j + 1] == a[i + 1][j] + a[i][j + 1]);    while (T --) {        x = read(); y = read(); k = read();        if (good[x + k - 2][y + k - 2] - good[x - 1][y + k - 2] - good[x + k - 2][y - 1] + good[x - 1][y - 1] == (k - 1) * (k - 1))            puts("Y");        else puts("N");    }    return 0;}//秦中花鸟已应阑，塞外风沙犹自寒。//    -- 王翰《凉州词二首》

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