• hdu 1575 Tr A


    题目连接

    http://acm.hdu.edu.cn/showproblem.php?pid=1575  

    Tr A

    Description

    A为一个方阵,则Tr A表示A的迹(就是主对角线上各项的和),现要求Tr(A^k)%9973。

    Input

    数据的第一行是一个T,表示有T组数据。
    每组数据的第一行有n(2 <= n <= 10)和k(2 <= k < 10^9)两个数据。接下来有n行,每行有n个数据,每个数据的范围是[0,9],表示方阵A的内容。

    Output

    对应每组数据,输出Tr(A^k)%9973。

    Sample Input

    2
    2 2
    1 0
    0 1
    3 99999999
    1 2 3
    4 5 6
    7 8 9

    Sample Output

    2
    2686

    #include<algorithm>
    #include<iostream>
    #include<cstdlib>
    #include<cstring>
    #include<cstdio>
    #include<vector>
    #include<map>
    using std::map;
    using std::min;
    using std::find;
    using std::pair;
    using std::vector;
    using std::multimap;
    #define pb(e) push_back(e)
    #define sz(c) (int)(c).size()
    #define mp(a, b) make_pair(a, b)
    #define all(c) (c).begin(), (c).end()
    #define iter(c) __typeof((c).begin())
    #define cls(arr, val) memset(arr, val, sizeof(arr))
    #define cpresent(c, e) (find(all(c), (e)) != (c).end())
    #define rep(i, n) for(int i = 0; i < (int)n; i++)
    #define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
    const int N = 1000007;
    const int M = 9973;
    typedef unsigned long long ull;
    struct Matrix {
        typedef vector<ull> vec;
        typedef vector<vec> mat;
        inline mat mul(mat &A, mat &B) {
            mat C(sz(A), vec(sz(B[0])));
            rep(i, sz(A)) {
                rep(k, sz(B)) {
                    rep(j, sz(B[0])) {
                        C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % M;
                    }
                }
            }
            return C;
        }
        inline mat pow(mat &A, int n) {
            mat B(sz(A), vec(sz(A[0])));
            rep(i, sz(A)) B[i][i] = 1;
            while(n) {
                if(n & 1) B = mul(B, A);
                A = mul(A, A);
                n >>= 1;
            }
            return B;
        }
        inline void solve(int n, int k) {
            ull v, ans = 0;
            mat ret(n, vec(n));
            rep(i, n) {
                rep(j, n) {
                    scanf("%lld", &v);
                    v %= M;
                    ret[i][j] = v;
                }
            }
            ret = pow(ret, k);
            rep(i, n) {
                ans += ret[i][i];
                ans %= M;
            }
            printf("%lld
    ", ans);
        }
    }go;
    int main() {
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w+", stdout);
    #endif
        int t, n, k;
        scanf("%d", &t);
        while(t--) {
            scanf("%d %d", &n, &k);
            go.solve(n, k);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/GadyPu/p/4792736.html
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