• UVA


                      Yet another Number Sequence

    Let’s define another number sequence, given by the following function:
    f(0) = a
    f(1) = b
    f(n) = f(n − 1) + f(n − 2), n > 1
    When a = 0 and b = 1, this sequence gives the Fibonacci Sequence. Changing the values of a and
    b, you can get many different sequences. Given the values of a, b, you have to find the last m digits of
    f(n).
    Input
    The first line gives the number of test cases, which is less than 10001. Each test case consists of a
    single line containing the integers a b n m. The values of a and b range in [0,100], value of n ranges in
    [0,1000000000] and value of m ranges in [1,4].
    Output
    For each test case, print the last m digits of f(n). However, you should NOT print any leading zero.
    Sample Input
    4
    0 1 11 3
    0 1 42 4
    0 1 22 4
    0 1 21 4
    Sample Output
    89
    4296
    7711
    946

    题意:

     给你 f[0],f[1] 分别为A,B求F[n] % (10^m)

    题解:

      n有点大,矩阵快速幂

    #include <iostream>
    #include <cstdio>
    #include <cmath>
    #include <cstring>
    #include <algorithm>
    using namespace std ;
    typedef long long ll;
    
    const int  N = 100000 + 10;
    const int mod = 1e9 + 7;
    const int M[55] = {1, 10, 100, 1000, 10000};
    struct Matrix {
        ll mat[2][2];
    }U,F,L;
    ll MOD;
    Matrix multi (Matrix a, Matrix b) {
        Matrix ans;
        for(int i = 0; i < 2; i++) {
            for(int j = 0; j < 2; j++) {
                ans.mat[i][j] = 0;
                for(int k = 0; k < 2; k++)
                    ans.mat[i][j] += a.mat[i][k] * b.mat[k][j];
                ans.mat[i][j] %= MOD;
            }
        }
        return ans;
    }
    ll a,b,m;
    Matrix powss(ll n) {
       Matrix ans = L,p = U;
       while(n) {
        if(n&1) ans = multi(p,ans);
        n >>= 1;
        p = multi(p,p);
       }
        return ans;
    }
    int main() {
    
        int T;
        scanf("%d",&T);
        while(T--) {
                ll n;
            scanf("%lld%lld%lld%lld",&a,&b,&n,&m);
            U = {1,1,1,0};
            L = {b,0,a,0};
            MOD = M[m];
            Matrix ans = powss(n);
            printf("%lld
    ",ans.mat[1][0]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/zxhl/p/5148580.html
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