HDU 5690 All X

矩阵快速幂

a[i]=a[i-1]*10+x; 问a[m]%k是否等于c

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<algorithm>
using namespace std;

long long x,m,ans,MOD;

struct Matrix
{
    long long A[5][5];
    int R, C;
    Matrix operator*(Matrix b);
};

Matrix X, Y, Z;

Matrix Matrix::operator*(Matrix b)
{
    Matrix c;
    memset(c.A, 0, sizeof(c.A));
    int i, j, k;
    for (i = 1; i <= R; i++)
        for (j = 1; j <= b.C; j++)
            for (k = 1; k <= C; k++)
                c.A[i][j] = (c.A[i][j] + (A[i][k] * b.A[k][j]) % MOD) % MOD;
    c.R = R; c.C = b.C;
    return c;
}

void init()
{
    memset(X.A, 0, sizeof X.A);
    memset(Y.A, 0, sizeof Y.A);
    memset(Z.A, 0, sizeof Z.A);

    Y.R = 3; Y.C = 3;
    for (int i = 1; i <= 3; i++) Y.A[i][i] = 1;

    X.R = 3; X.C = 3;
    X.A[1][1]=10;
    X.A[2][2]=10;
    X.A[3][1]=1; X.A[3][2]=1; X.A[3][3]=1;

    Z.R = 1; Z.C = 3;
    Z.A[1][1]=0; Z.A[1][2]=x; Z.A[1][3]=x;

}

void read()
{
    scanf("%lld%lld%lld%lld",&x,&m,&MOD,&ans);
}

void work()
{
    while (m)
    {
        if (m % 2 == 1) Y = Y*X;
        m = m >> 1;
        X = X*X;
    }
    Z = Z*Y;

    if(Z.A[1][1]==ans) printf("Yes
");
    else printf("No
");

}

int main()
{
    int Case=1;
    int T;
    scanf("%d", &T);
    while (T--)
    {
        read();
        init();
        printf("Case #%d:
",Case++);
        work();
    }
    return 0;
}
原文地址:https://www.cnblogs.com/zufezzt/p/5519077.html