2015
07-16

# Homework

GS is suffering from tons of boring math assignment. He find it make him tired and impatient so he asks you to finish his assignment in hope that he could hang out in many places of interest and enjoy his life.
In this assignment, you’re asked to solve the following problem:
Given a recurrent function

and boundary values

You should solve for f[n].
What a easy problems! Wait for a moment, you see a few lines in the last paragraph. It reads as follows: To make the problem a little hard, you are now informed that, at some special values of n (there are q such values, namely n1, n2, . . . , nq), the recurrent formula changes into something else, which means for the kth such value nk, the recurrent formula changes into

Still an easy problem, isn’t it?
Since f[n] may be quite large, you just need to output f[n] module 109 + 7.

There are several test cases.
For each test case, the first line contains three integers n (m < n ≤ 109), m (1 ≤ m ≤ 100), q (0 ≤ q ≤ 100). The second line contains m integers, namely f[1], f[2], . . . , f[m].
The following line contains several integers, first comes t (t ≤ 100), then t integers namely c[1], c[2], . . . , c[t].
The following q lines describe q special cases of the recurrent formula, each containing several integers, namely nk, tk (tk ≤ 100, tk < nk), ck[1], ck[2], . . . , ck[tk], as mentioned earlier. It is satisfied that ni != nj if i != j.
All integers are non-negative. Unless specified, all integers are not greater than 109.
Input is terminated by EOF.
You might assume that all given data is correct.

There are several test cases.
For each test case, the first line contains three integers n (m < n ≤ 109), m (1 ≤ m ≤ 100), q (0 ≤ q ≤ 100). The second line contains m integers, namely f[1], f[2], . . . , f[m].
The following line contains several integers, first comes t (t ≤ 100), then t integers namely c[1], c[2], . . . , c[t].
The following q lines describe q special cases of the recurrent formula, each containing several integers, namely nk, tk (tk ≤ 100, tk < nk), ck[1], ck[2], . . . , ck[tk], as mentioned earlier. It is satisfied that ni != nj if i != j.
All integers are non-negative. Unless specified, all integers are not greater than 109.
Input is terminated by EOF.
You might assume that all given data is correct.

7 5 0
1 1 2 3 5
2 1 1
10 5 1
1 1 2 3 5
2 1 1
10 2 1 2

Case 1: 13
Case 2: 76
Hint
In the first sample, you are to solve for f[7] where f[n] = f[n−1]+f[n−2] and f[1] = 1,
f[2] = 1, f[3] = 2, f[4] = 3, f[5] = 5.
In the second example, you are to solve for f[10] where f[n] = f[n − 1] + f[n − 2] and
f[1] = 1, f[2] = 1, f[3] = 2, f[4] = 3, f[5] = 5, as well as specially f[10] = f[9] + 2*f[8].


http://acm.hdu.edu.cn/showproblem.php?pid=4471

D =

 c1 c2 “ c[h-1] c[h] 1 0 “ 0 0 0 1 “ 0 0 0 0 0 0 0 0 1 0

V[x] =

 f[x] f[x-1]   f[x-h+1]

http://www.07net01.com/program/547544.html

 #include<cstdio>
#include<algorithm>
using namespace std;
const int N =102;
const int mod = 1000000007;
int h;//计算f[x]时最多和前面h个数有关
struct matrix
{
int row,col;
int m[N][N];
void init(int row,int col)
{
this->row = row;
this->col = col;
for(int i=0; i<=row; ++i)
for(int j=0; j<=col; ++j)
m[i][j] = 0;
}
} A,pm[33],ans;

matrix operator*(const matrix & a,const matrix& b)
{
matrix res;
res.init(a.row,b.col);
for(int k=1; k<=a.col; ++k)
{
for(int i=1; i<= res.row; ++i)
{
if(a.m[i][k] == 0 ) continue;
for(int j = 1; j<=res.col; ++j)
{
if(b.m[k][j] == 0 ) continue;
res.m[i][j] = (1LL *a.m[i][k]*b.m[k][j] + res.m[i][j])%mod;
}
}
}
return res;
}

void cal(int x)
{
for(int i=0; i<=31; ++i)
if(x & (1<<i) ) ans = pm[i]*ans;
}
void getPm()
{
pm[0] = A;
for(int i=1; i<=31; ++i)
pm[i] = pm[i-1]*pm[i-1];
}
struct sp//特殊点
{
int nk,tk;//nk为点的位置，tk为计算nk时和前面tk个数有关
int ck[N];
bool operator<(const sp & o)const//按照nk排序
{
return nk<o.nk;
}
} p[N];
int main()
{
//    freopen("in.txt","r",stdin);
int n,m,q,t,f[N],c[N],kase=0;
while(~scanf("%d%d%d",&n,&m,&q))
{
for(int i=m; i>0; --i)   scanf("%d",&f[i]);
scanf("%d",&t);
h =t;
for(int i=1; i<=t; ++i)  scanf("%d",&c[i]);
for(int i=0; i<q; ++i)
{
scanf("%d%d",&p[i].nk,&p[i].tk);
if(p[i].tk > h) h = p[i].tk;
for(int j=1; j<=p[i].tk; ++j) scanf("%d",&p[i].ck[j]);
}
sort(p,p+q);
A.init(h,h);
for(int i=1; i<=t; ++i) A.m[1][i] = c[i];
for(int i=2; i<=h; ++i)  A.m[i][i-1] = 1;
getPm();
ans.init(h,1);
for(int i = m; i > 0; --i)   ans.m[i][1] = f[i];
int last=m;
for(int i=0; i<q; ++i)
{
if( p[i].nk <=last ||  p[i].nk >n ) continue;
cal( p[i].nk-last-1);
last =  p[i].nk;
for(int j=1; j<=p[i].tk; ++j)  A.m[1][j] = p[i].ck[j];
for(int j=p[i].tk+1; j<=h; ++j) A.m[1][j] = 0;
ans  =A*ans;
}
cal(n-last);
printf("Case %d: %d\n",++kase,ans.m[1][1]);
}
return 0;
}


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