hdoj-1757-A Simple Math Problem【矩阵的快速幂】

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3465 Accepted Submission(s): 2087



[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
For each case, output f(k) % m in one line.

[align=left]Sample Input[/align]

10 9999
1 1 1 1 1 1 1 1 1 1
20 500
1 0 1 0 1 0 1 0 1 0

[align=left]Sample Output[/align]

45
104

[align=left]Author[/align]
linle

[align=left]Source[/align]
2007省赛集训队练习赛（6）_linle专场

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#include<stdio.h>
#include<string.h>
struct matrix{
int m[10][10];
}ans,base;
int MOD;
matrix multi(matrix a,matrix b){
matrix temp;
for(int i=0;i<10;++i){
for(int j=0;j<10;++j){
temp.m[i][j]=0;
for(int k=0;k<10;++k)
temp.m[i][j]=(temp.m[i][j]+a.m[i][k]*b.m[k][j])%MOD;
}
}
return temp;
}
void output(matrix a){
for(int i=0;i<10;++i){
for(int j=0;j<10;++j)
printf("%d ",a.m[i][j]);
printf("\n");
}
printf("**************\n");
}
matrix fast_mod(int n) {//求base的 n 次方
//为base 赋初值
memset(base.m,0,sizeof(base.m));
for(int i=0;i<10;++i){
scanf("%d",&base.m[0][i]);
}
for(int i=1;i<10;++i)	base.m[i][i-1]=1;
// 将 矩阵ans 初始化为单位矩阵
//output(base);
memset(ans.m,0,sizeof(ans.m));
for(int i=0;i<10;++i) ans.m[i][i]=1;
//	output(ans);
while(n){
if(n & 1)  ans=multi(ans,base);

base=multi(base,base);
n>>=1;
}
return ans;
}
int main(){
int k,m;
while(~scanf("%d%d",&k,&m)){
MOD=m;
fast_mod(k-9);
int i,j,ts=0;
for(i=0,j=9;i<9;++i,--j){
ts=(ts+ans.m[0][i]*j)%MOD;
}
printf("%d\n",ts);
}
return 0;
}