hdoj 4990 Reading comprehension（矩阵快速幂）

找到递推式，接下来构造矩阵就可以了。

还要注意用long long存矩阵

代码：

#include<bits/stdc++.h>using namespace std;typedef long long ll;ll m;struct node{    ll s[3][3];    node() {}    node(ll a, ll b, ll c, ll d, ll e, ll f, ll g, ll h, ll i)    {        s[0][0] = a;        s[0][1] = b;        s[0][2] = c;        s[1][0] = d;        s[1][1] = e;        s[1][2] = f;        s[2][0] = g;        s[2][1] = h;        s[2][2] = i;    }};node mul(node a, node b){    node t;    memset(t.s, 0, sizeof(t.s));    for(int i = 0; i < 3; i++)        for(int j = 0; j < 3; j++)            for(int k = 0; k < 3; k++)                t.s[i][j] = (t.s[i][j]+a.s[i][k]*b.s[k][j])%m;    return t;}node mt_pow(node p, int n){    node q;    memset(q.s, 0, sizeof(q.s));    q.s[0][0] = q.s[1][1] = q.s[2][2] = 1;    while(n)    {        if(n&1) q = mul(p, q);        p = mul(p, p);        n /= 2;    }    return q;}int main(void){    ll n;    while(cin >> n >> m)    {        if(n == 1) printf("%d\n", 1%m);        else if(n == 2) printf("%d\n", 2%m);        else        {            node base = node(1, 2, 1, 1, 0, 0, 0, 0, 1);            node ans = mt_pow(base, n-2);            printf("%lld\n", (ans.s[0][0]*2+ans.s[0][1]+ans.s[0][2])%m);        }    }    return 0;}

Reading comprehension

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1520    Accepted Submission(s): 611

Problem Description

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

1 103 100

 

Sample Output

15

 

Source

BestCoder Round #8

 

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