uva 11024 - Circular Lock(數學)

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題目連結：uva 11024 - Circular Lock

題目大意；有個2*2的矩陣，給定p，s，P為p數組中所有元素的最大公約數。s為2*2矩陣的初始狀態，每次可以選擇一行或是一列同時加1，最終使得sij%P=0

解題思路：gij為aij還需要多少可以是P的倍數，判斷g11?g12?g21+g22是P的倍數即可。

/******************** * 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;}