Hihocoder-1286 子矩阵求和

来源：互联网发布：淘宝秒杀助手官方下载编辑：程序博客网时间：2024/06/06 04:19

解题思路：

看到这个题目的时候是很懵逼的= =矩阵是无限的

但是其实没那么刚，只需要巧妙的转换下就可以得到结果。

对于矩阵：

1 1 1 1 1 1 1 11 2 2 2 2 2 2 21 2 3 3 3 3 3 31 2 3 4 4 4 4 41 2 3 4 5 5 5 51 2 3 4 5 6 6 61 2 3 4 5 6 7 71 2 3 4 5 6 7 8

对于这样的矩阵，我们能看到两个规律：

用s[i][j]表示左上角左边为(i, j)的子矩阵和

1.对于要求的N*M的子矩阵，当行数超过M后，s[i][j] = s[i+1][j](i >= M),当列数超过N的时候，s[i][j] = s[i][j+1](j >= N)

2.s[i][j] + x * N * M = s[i+x][j+x]

所以这题的做法就出来了，只需要枚举第一行和第一列，对于每个位置，求(s[i][j] + x*N*M) mod k = 0的最小x，这个时候很自然的想到扩展欧几里得算法。所以求一发就解出来了。

代码：

#include <set>#include <map>#include <cmath>#include <queue>#include <stack>#include <cstdio>#include <vector>#include <string>#include <cstring>#include <cstdlib>#include <iostream>#include <algorithm>#include <functional>using namespace std;typedef long long LL;const LL inf = 1e18 + 5;LL x, y, ans_x, ans_y;LL extend_gcd(LL a, LL b, LL& x, LL &y) {    if ( b == 0 ) { x = 1; y = 0; return a; }    LL ans = extend_gcd( b, a % b, y, x );    y -= a / b * x;    return ans;}LL solve( LL n, LL m, LL cur, LL sum ) {    LL i = n + cur - 1;    if ( i > m ) sum += (m + 1) * m / 2;    else{        sum += (i + 1) * i / 2;        sum += i * (m - i);    }    if ( cur - 1 > m ) sum -= (m + 1) * m / 2;    else {        sum -= cur * (cur - 1) / 2;        sum -= (cur - 1) * (m - cur + 1);    }    return sum;}bool judge( LL a, LL b, LL c, LL d, LL base1, LL base2 ) {    if ( (c % d) == 0 ) {        a /= d; b /= d; c /= d;        LL tmp = ((x % b) * (c % b)) % b;        while(tmp < 0) tmp += b >= 0 ? b : -b;        LL tt = tmp * 2 + base1 + base2;        if ( tt < ans_x + ans_y || (tt == ans_x + ans_y && tmp + base1 < ans_x) || (tt == ans_x + ans_y && tmp + base1 == ans_x && tmp + base2 < ans_y) ) {            ans_x = base1 + tmp;            ans_y = base2 + tmp;        }        return true;    }    return false;}int main() {    LL q, n, m, k;    cin >> q;    while(q--) {        LL init = 0;        cin >> n >> m >> k;        for ( LL i = 1; i <= n; ++i ) {            if ( i > m ) init += (m + 1) * m / 2;            else {                init += (i + 1) * i / 2;                init += i * (m - i);            }        }        ans_x = ans_y = inf;        bool flag = false;        LL A = -n * m, B = k, C = init;        LL D = extend_gcd(A, B, x, y);        flag |= judge(A, B, C, D, 1, 1);        LL sum = init;        for ( LL i = 2; i <= n; ++i ) {            sum = solve(m, n, i, sum);            flag |= judge(A, B, sum, D, 1, i);        }        sum = init;        for ( LL i = 2; i <= m; ++i ) {            sum = solve(n, m, i, sum);            flag |= judge(A, B, sum, D, i, 1);        }        if ( !flag ) cout << "-1" << endl;        else cout << ans_x << " " << ans_y << endl;    }    return 0;}

0 0