uva 11024 - Circular Lock(数学)

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题目链接：uva 11024 - Circular Lock

题目大意；有个2*2的矩阵，给定p，s，P为p数组中所有元素的最大公约数。s为2*2矩阵的初始状态，每次可以选择一行或是一列同时加1，最终使得sij%P=0

解题思路：gij为aij还需要多少可以是P的倍数，判断g11−g12−g21+g22是P的倍数即可。

/******************** * A + C = a + k1 * p * B + C = b + k2 * p * A + D = c + k3 * p * B + D = d + k4 * p * * a - b - c + d + (k1 - k2 - k3 + k4) * p *  = 0;********************/#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 5;int s[maxn][maxn], p[maxn][maxn];inline int gcd (int a, int b) {    return b == 0 ? a : gcd(b, a % b);}int main () {    int cas;    scanf("%d", &cas);    while (cas--) {        for (int i  = 1; i <= 2; i++)            scanf("%d %d %d %d", &s[i][1], &s[i][2], &p[i][1], &p[i][2]);        int P = p[1][1];        for (int i = 1; i <= 2; i++)            for (int j = 1; j <= 2; j++)                P = gcd(P, p[i][j]);        int sum = 0;        for (int i = 1; i <= 2; i++) {            for (int j = 1; j <= 2; j++) {                s[i][j] = P - s[i][j] % P;                if (i == j)                    sum += s[i][j];                else                    sum -= s[i][j];            }        }        printf("%s\n", sum % P == 0 ? "Yes" : "No");    }    return 0;}

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