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hdu 1757 A Simple Math Problem(矩阵快速幂)
阅读量:136 次
发布时间:2019-02-27

本文共 2121 字,大约阅读时间需要 7 分钟。

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?????????????????????????????????????????????????????????????

1. ???????

???????????????????????????????????????? $A$ ? $n$ ??????? $A^n = A \cdot A \cdot \ldots \cdot A$?? $n$ ? $A$??????????? $O(n)$ ??????????????????????????????????????? $O(\log n)$??????????

2. ????

???????????

#include 
using namespace std;#define mod(x) ((x) % m)int k, m;struct mat { int d[10][10]; mat operator*(const mat x) { mat ret; int tmp; for (int i = 0; i < 10; i++) { for (int j = 0; j < 10; j++) { tmp = 0; for (int k = 0; k < 10; k++) { tmp = mod(tmp + d[i][k] * x.d[k][j]); } ret.d[i][j] = tmp; } } return ret; } void init_unit() { for (int i = 0; i < 10; i++) for (int j = 0; j < 10; j++) d[i][j] = (i == j) ? 1 : 0; } void init() { for (int i = 0; i < 10; i++) for (int j = 0; j < 10; j++) d[i][j] = (i == j + 1) ? 1 : 0; } void set(int i, int v) { d[9 - i][9] = v; }};mat fastPow(mat base, int pow) { mat res; res.init_unit(); while (pow) { if (pow & 1) res = res * base; base = base * base; pow >>= 1; } return res;}int main() { int tmp; while (scanf("%d%d", &k, &m) == 2) { if (k < 10) { printf("%d\n", k % m); continue; } a.init(); for (int i = 0; i < 10; i++) scanf("%d", &tmp), a.set(i, tmp); a = fastPow(a, k - 9); int ans = 0; for (int i = 0; i < 10; i++) ans = mod(ans + a.d[i][9] * i); printf("%d\n", ans); } return 0;}

3. ????

  • ?????????

    ??? mat ????? 10x10 ?????????????????????????????????????

  • ?????

    • init_unit() ????????????? 1???? 0??
    • init() ????????????????????? 1???? 0?
    • set() ????????????????
  • ?????

    fastPow ???????????????????????????????????????????????????

  • ?????

    ?????? k ? m?????? a????????????? a ???????????????

  • 4. ??

    ???????????????????????????????????????????????????????????????????????????

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