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  • 矩阵快速幂 UVA 10870 Recurrences

    题目传送门

    题意:f(n) = a1f(n − 1) + a2f(n − 2) + a3f(n − 3) + . . . + adf(n − d), for n > d,求f (n) % m。训练指南的题目

    分析:令:.则

    #include <bits/stdc++.h>
    
    int d, n, m;
    int a[16], f[16];
    
    struct Mat {
        int m[17][17];
        int row, col;
        Mat() {
            //row = col = 16;
            memset (m, 0, sizeof (m));
        }
        void init(int sz) {
            row = col = sz;
            for (int i=1; i<row; ++i) {
                m[i][i+1] = 1;
            }
            int c = sz - 1;
            for (int i=2; i<=col; ++i) {
                m[sz][i] = a[c--];
            }
        }
        void change(int sz) {
            row = col = sz;
            for (int i=1; i<=sz; ++i) {
                m[i][i] = 1;
            }
        }
    };
    
    Mat operator * (const Mat &a, const Mat &b) {
        Mat ret;
        ret.row = a.row; ret.col = b.col;
        for (int i=1; i<=a.row; ++i) {
            for (int j=1; j<=b.col; ++j) {
                for (int k=1; k<=a.col; ++k) {
                    int &r = ret.m[i][j];
                    r = (r + 1ll * a.m[i][k] * b.m[k][j]) % m;
                }
            }
        }
        return ret;
    }
    
    Mat operator ^ (Mat x, int n) {
        Mat ret; ret.change (d+1);
        while (n) {
            if (n & 1) {
                ret = ret * x;
            }
            x = x * x; n >>= 1;
        }
        return ret;
    }
    
    //Running_Time
    int main() {
        while (scanf ("%d%d%d", &d, &n, &m) == 3) {
            if (!d && !n && !m) {
                break;
            }
            for (int i=1; i<=d; ++i) {
                scanf ("%d", a+i);
            }
            for (int i=1; i<=d; ++i) {
                scanf ("%d", f+i);
            } 
            if (n <= d) {
                printf ("%d
    ", f[n] % m);
            } else {
                Mat ans, Fd;
                ans.init (d + 1);
                ans = ans ^ (n - d);
                
                Fd.row = d + 1; Fd.col = 1;
                for (int i=2; i<=d+1; ++i) {
                    Fd.m[i][1] = f[i-1];
                }
    
                ans = ans * Fd;
                printf ("%d
    ", ans.m[d+1][1]);
            }
        }
    
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Running-Time/p/5365100.html
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