HDU 2604 Queuing（矩阵快速幂）

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题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=2604

思路：刚开始写DP错了，看别人的题解改了还是错，直接学快速幂A了，学了一招，第二位有2维时，可以开2个一维数组，比较容易理解

AC代码：

错误代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <cstring>#include <climits>#include <cmath>#include <cctype>const int inf = 0x3f3f3f3f;//1061109567typedef long long ll;using namespace std;int dp[1000010][2];int main(){    int l,m;    while(scanf("%d%d",&l,&m) != EOF)    {        dp[1][0] = 1;        dp[1][1] = 1;        dp[2][0] = 2;        dp[2][1] = 2;        for(int i=3; i<=l; i++)        {            //dp[i][0] = dp[i-2][1];            dp[i][0] = (2 * dp[i-2][1] - dp[i-3][0]) % m;            dp[i][1] = (dp[i-1][0] + dp[i-1][1]) % m;        }        printf("%d\n",(dp[l][0] + dp[l][1])%m);    }    return 0;}

错误代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <cstring>#include <climits>#include <cmath>#include <cctype>const int inf = 0x3f3f3f3f;//1061109567typedef long long ll;using namespace std;int f[1000010];int m[1000010];int main(){    int l,k;    while(scanf("%d%d",&l,&k) != EOF)    {        f[1] = 1;        m[1] = 1;        f[2] = 2;        m[2] = 2;        for(int i=3; i<=l; i++)        {            f[i] = (m[i-2] * 2 - f[i-3]) % k;            m[i] = (m[i-1] + f[i-1]) % k;        }        printf("%d\n",(f[l] + m[l])%k);    }    return 0;}

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