0
点赞
收藏
分享

微信扫一扫

【HDU - 4990】 Reading comprehension (构造+矩阵快速幂)

有点d伤 2022-06-15 阅读 46

题干:

Read the program below carefully then answer the question. 
#pragma comment(linker, "/STACK:1024000000,1024000000") 
#include <cstdio> 
#include<iostream> 
#include <cstring> 
#include <cmath> 
#include <algorithm> 
#include<vector> 

const int MAX=100000*2; 
const int INF=1e9; 

int main() 

  int n,m,ans,i; 
  while(scanf("%d%d",&n,&m)!=EOF) 
  { 
    ans=0; 
    for(i=1;i<=n;i++) 
    { 
      if(i&1)ans=(ans*2+1)%m; 
      else ans=ans*2%m; 
    } 
    printf("%d\n",ans); 
  } 
  return 0; 
}

Input

Multi test cases,each line will contain two integers n and m. Process to end of file. 
[Technical Specification] 
1<=n, m <= 1000000000

Output

For each case,output an integer,represents the output of above program.

Sample Input


1 10 3 100

Sample Output


1 5

解题报告:

      根据题干找规律,发现f(n) = f(n-1) + 2 * f(n-2) +1

AC代码:

#pragma comment(linker, "/STACK:1024000000,1024000000") 
#include <cstdio>
#include<iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#define ll long long
struct Matrix {
ll arr[4][4];
}unit,trans;
ll n,m,ans;
Matrix mul( Matrix a,Matrix b,ll mod) {
Matrix c;
for(int i = 1; i<=3; i++) {
for(int j = 1; j<=3; j++) {
c.arr[i][j] = 0;
for(int k = 1; k<=3; k++) {
// if(a.arr[i][k] && b.arr[k][j])
c.arr[i][j] = (c.arr[i][j] + a.arr[i][k]*b.arr[k][j])%mod;
}
}
}
return c;
}
Matrix q_pow(Matrix a, ll k,ll mod) {
Matrix ans;
ans = unit;
while(k) {
if(k&1) {
ans=mul(ans,a,m);
}
k>>=1;
a=mul(a,a,m);
}
return ans;
}
void init() {
memset(unit.arr,0,sizeof(unit.arr) );
for(int i = 1; i<=3; i++) {
unit.arr[i][i] = 1;
trans.arr[i][1] = trans.arr[i][2] = trans.arr[i][3] = 0;
}
trans.arr[1][1]=trans.arr[1][3] =trans.arr[2][1]=trans.arr[3][3] = 1;
trans.arr[1][2] = 2;
}
int main()
{
Matrix ans;
ll sum=0;
while(scanf("%lld%lld",&n,&m)!=EOF) {
sum = 0;
init();
if(n==1) {
printf("%lld\n",1%m);
continue;
}
else if(n == 2) {
printf("%lld\n",2%m);continue;
}
ans = q_pow(trans,n-2,m);
sum = ans.arr[1][1]*2 + ans.arr[1][2]*1 + ans.arr[1][3];
printf("%lld\n",sum%m);
}
return 0;
}

总结:

      这题坑很多啊,总是巧妙的跳了进去。

      1.全局变量ll的n和m,结果输入用%d这个题的Mul函数 不是i<=n了!而是i<=3即可,i<=n就错了!  此题和​​HDU - 5015​​不一样,不是矩阵是个不确定的矩阵!所以这里是小于等于3!!即

      2.输出的时候n=1和n=2需要特判一下,因为你传参是n-2。即:这种地方一定要小心!你的函数是有使用条件的,这一点不仅适用于矩阵快速幂,还有很多其他的地方。

     3.输出n==2的时候,别忘了也需要取模!!

    4.最后的输出 也需要取模!


举报

相关推荐

0 条评论