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
    param[:]=param-lr*param.grad/batch_size的理解
    查看>>
    spring mvc excludePathPatterns失效 如何解决spring拦截器失效 excludePathPatterns忽略失效 拦截器失效 spring免验证拦截器不起作用
    查看>>
    Spring Cloud 之注册中心 EurekaServerAutoConfiguration源码分析
    查看>>
    Parrot OS 6.2 重磅发布!推出全新 Docker 容器启动器
    查看>>
    Parrot OS 6.3 发布!全面提升安全性,新增先进工具,带来更高性能
    查看>>
    ParseChat应用源码ios版
    查看>>
    Part 2异常和错误
    查看>>
    Pascal Script
    查看>>
    Spring Boot集成Redis实现keyspace监听 | Spring Cloud 34
    查看>>
    Spring Boot中的自定义事件详解与实战
    查看>>
    Passport 密码模式
    查看>>
    Spring Boot(七十六):集成Redisson实现布隆过滤器(Bloom Filter)
    查看>>
    passwd命令限制用户密码到期时间
    查看>>
    Spring Boot 动态加载jar包,动态配置太强了!
    查看>>
    Spring @Async执行异步方法的简单使用
    查看>>
    PAT (Basic Level) Practice 乙级1021-1030
    查看>>
    PAT (Basic Level) Practice 乙级1031-1040
    查看>>
    PAT (Basic Level) Practice 乙级1041-1045
    查看>>
    SparkSql的元数据
    查看>>
    PAT (Basic Level) Practice 乙级1051-1055
    查看>>