博客
关于我
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。

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

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

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. ??

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

    转载地址:http://dwib.baihongyu.com/

    你可能感兴趣的文章
    poj 2965 The Pilots Brothers' refrigerator-1
    查看>>
    poj 3026( Borg Maze BFS + Prim)
    查看>>
    POJ 3041 - 最大二分匹配
    查看>>
    POJ 3041 Asteroids(二分匹配模板题)
    查看>>
    Qt笔记——标准文件对话框QFileDialog
    查看>>
    poj 3083 Children of the Candy Corn
    查看>>
    POJ 3083 Children of the Candy Corn 解题报告
    查看>>
    POJ 3253 Fence Repair C++ STL multiset 可解 (同51nod 1117 聪明的木匠)
    查看>>
    Qt笔记——控件总结
    查看>>
    poj 3262 Protecting the Flowers 贪心
    查看>>
    poj 3264(简单线段树)
    查看>>
    Qt笔记——布局管理三件套分割窗口、停靠窗口和堆栈窗口
    查看>>
    poj 3277 线段树
    查看>>
    POJ 3349 Snowflake Snow Snowflakes
    查看>>
    POJ 3411 DFS
    查看>>
    poj 3422 Kaka's Matrix Travels (费用流 + 拆点)
    查看>>
    Qt笔记——官方文档全局定义(二)Functions函数
    查看>>
    POJ 3468 A Simple Problem with Integers
    查看>>
    poj 3468 A Simple Problem with Integers 降维线段树
    查看>>
    poj 3468 A Simple Problem with Integers(线段树 插线问线)
    查看>>