Description
Lele now is thinking about a simple function f(x).
If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .
Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.
If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .
Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.
Input
The problem contains mutiple test cases.Please process to the end of file.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.
Output
For each case, output f(k) % m in one line.
Sample Input
10 9999
1 1 1 1 1 1 1 1 1 1
20 500
1 0 1 0 1 0 1 0 1 0
Sample Output
45
104
Please process to the end of file
沒注意這句話，WA了好幾次。。
#include <iostream>
#include <cstdio>
#include <cstring>
int n , m;
int A[12];
struct Node
{
int m[12][12];
int x;
int y;
}pt , st , dt;
Node Pow(Node a, Node b)
{
memset(dt.m , 0 , sizeof(dt.m));
dt.x = a.x;
dt.y = b.y;
for(int i = 1 ; i <= a.x; i++)
{
for(int k = 1; k <= a.y; k++)
{
if(a.m[i][k] == 0)
continue;
{
for(int j = 1; j <= b.y; j++)
{
dt.m[i][j] = ((dt.m[i][j] + a.m[i][k] * b.m[k][j]) % m) % m;
}
}
}
}
return dt;
}
void quickpow(int n)
{
while(n)
{
if(n % 2 == 1)
pt = Pow(st , pt);
n = n / 2;
st = Pow(st , st);
}
}
int main()
{
while(~scanf("%d %d",&n , &m))
{
memset(pt.m , 0 ,sizeof(pt.m));
memset(st.m , 0 , sizeof(st.m));
pt.x = pt.y = 10;
st.x = st.y = 10; //矩陣的行列長度
for(int i = 0; i <= 9; i++)
scanf("%d",&st.m[i+1][1]);
for(int i = 1; i <= 10; i++)
pt.m[i][i] = 1;
for(int i = 2; i <= 10; i++)
{
st.m[i-1][i] = 1;//寫成了st.m[i][i-1] 檢查了n多次 才找到
}
if(n < 9)
printf("%d\n",n%m);
else
{
quickpow(n-9);
int sum;
sum = 9*pt.m[1][1]+8*pt.m[2][1]+7*pt.m[3][1]+6*pt.m[4][1]+5*pt.m[5][1]+4*pt.m[6][1]+3*pt.m[7][1]+2*pt.m[8][1]+1*pt.m[9][1];
printf("%d\n",sum%m);
}
}
return 0;
}