題干：

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.?

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.?

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

?

題目大意：

? 告訴你f(x)的運算式，輸入k和m，輸出f(k)%m

解題報告：

? 好久不寫博客了。。來復習一波矩陣快速冪吧、、

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e5 + 5; ll mod,k; ll a[15]; struct Matrix {ll arr[15][15];//其實應該是11*11的 } unit; Matrix Mul(Matrix a,Matrix b) {Matrix c;for(int i = 1; i<=11; i++) {for(int j = 1; j<=11; j++) {c.arr[i][j]=0;for(int k = 1; k<=11; k++) {c.arr[i][j] = (c.arr[i][j] + a.arr[i][k]*b.arr[k][j])%mod;}}}return c; } Matrix qpow(Matrix a,ll k) {Matrix res = unit;while(k) {if(k&1) res = Mul(res,a);a=Mul(a,a);k>>=1;}return res; } int main() {for(int i = 1; i<=11; i++) {unit.arr[i][i]=1;}while(~scanf("%lld%lld",&k,&mod)) {for(int i = 1; i<=10; i++) scanf("%lld",a+i);Matrix trans;//INIT for(int i = 1; i<=11; i++) {for(int j = 1; j<=11; j++) {if(i == 1) trans.arr[i][j] = a[j];else {if(i-1 == j) trans.arr[i][j] = 1;else trans.arr[i][j]=0;}}}if(k <= 9) {printf("%lld\n",k);}else {Matrix ans = qpow(trans,k-9);ll out = 0;for(int i = 1; i<=10; i++) {out += ans.arr[1][i]*(10-i);out %= mod;}printf("%lld\n",out);}}return 0 ;}

?

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