NEU 1040 Count

最新推荐文章于 2024-11-22 15:04:43 发布

转载最新推荐文章于 2024-11-22 15:04:43 发布 · 105 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/xuwangzihao/p/4992574.html

本文介绍了一种使用快速幂优化的矩阵乘法方法来解决斐波那契数列求解问题，针对给定的V数量n和取余数p，计算并输出序列的组合数对p取余后的结果。通过实例输入输出展示了求解过程及应用。

1040: Count

时间限制: 1 Sec 内存限制: 128 MB
提交: 59 解决: 23
[提交][状态][讨论版]

题目描述

Many ACM team name may be very funny,such as "Complier_Error","VVVVV".Oh,wait for a minute here.

Is it "W W"+"V",or "W"+"V V V",or something others we can treat as?There are several ways we can treat this name "VVVVV" (5 'V's),as V V can be treat as a W.

For 5 'V's,our have 8 ways.They are:

V V V V V
V W W
W W V
V W V V
W V W
W V V V
V V W V
V V V W

The problem here is that for n 'V's,how many ways do we have to treat it?Because the answer may be too large, you should output the answer module by p.(If n is 0,then we have just one way.)

输入

There are multiple test cases. The first line of the input contains an integer M, meaning the number of the test cases.
For each test cases, there are two integers n and p in a single line.
You can assume that 0<=n<=2100000000, 0<p<=2009.

输出

For each test case, output the answer with case number in a single line.

样例输入

2
5 5
4 7

样例输出

3
5


题目大意：
就是说给你5个V的话“VVVVV”，可能有一部分V连在一起被看做W，问可以看出多少种序列，答案对p取余。
输入：
第一行一个M表示有M个测试数据，
第二行到第M+1行每行两个整数n，p，表示n个V，对p取余。
输出：
一个整数，要求如题。

菲波那切数列？？应该吧

好吧，就是斐波那契额，n的数目由n-1和n-2继承来，n比较大，所以矩阵乘法快速幂优化一下递推

 1 #include<cstdio>
 2 #include<cstring>
 3 int p;
 4 struct matrix
 5 {
 6     int a[2][2];
 7     matrix(matrix &p)
 8     {
 9         for(int i=0;i<2;i++)
10             for(int j=0;j<2;j++)
11                 this->a[i][j]=p.a[i][j];
12     }
13     matrix(int x)
14     {
15         for(int i=0;i<2;i++)
16             for(int j=0;j<2;j++)
17                 this->a[i][j]=x;
18     }
19     matrix()
20     {
21         memset(a,0,sizeof(a));
22         for(int j=0;j<2;j++)
23             this->a[j][j]=1;
24     }
25     matrix operator * (matrix &b)
26     {
27         matrix c;
28         for(int i=0;i<2;i++)
29             for(int j=0;j<2;j++)
30             {
31                 c.a[i][j]=0;
32                 for(int k=0;k<2;k++)
33                 {
34                     c.a[i][j]+=this->a[i][k]*b.a[k][j];
35                 }
36                 c.a[i][j]%=p;
37             }
38         return c;
39     }
40 };
41 matrix quickmult(matrix &a,int k)
42 {
43     matrix ans,temp(a);
44     while(k)
45     {
46         if(k%2)ans=ans*temp;
47         temp=temp*temp;
48         k/=2;
49     }
50     return ans;
51 }
52 int main()
53 {
54     int m,n;
55     scanf("%d",&m);
56     while(m--)
57     {
58         scanf("%d%d",&n,&p);
59         matrix ini,tra;//ini means initial matrix, tra means transform matrix
60         ini.a[0][0]=1;ini.a[0][1]=1;ini.a[1][0]=0;ini.a[1][1]=0;
61         tra.a[0][0]=0;tra.a[0][1]=1;tra.a[1][0]=1;tra.a[1][1]=1;
62         tra=quickmult(tra,n);
63         ini=ini*tra;
64         printf("%d\n",ini.a[0][0]);
65     }
66     return 0;
67 }

转载于:https://www.cnblogs.com/xuwangzihao/p/4992574.html