A Simple Math Problem
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5686 Accepted Submission(s): 3481
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.
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.
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <queue>
#include <stack>
#include <math.h>
#include <algorithm>
#include <map>
#include <vector>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 11;
const int n = 10;
ll mod, k;
struct Matrix
{
ll m[maxn][maxn];
};
Matrix multi(Matrix a, Matrix b)
{
Matrix c;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
c.m[i][j] = 0;
for (int k = 0; k < n; k++) {
c.m[i][j] += a.m[i][k] * b.m[k][j]%mod;
}
c.m[i][j] %= mod;
}
}
return c;
}
Matrix pow(Matrix a, ll nn)
{
Matrix ans;
memset(ans.m, 0, sizeof(ans.m));
for(int i = 0; i < n; i++) {
ans.m[i][i] = 1;
}
Matrix p = a;
while (nn) {
if (nn & 1)
ans = multi(ans, p);
p = multi(p, p);
nn /= 2;
}
return ans;
}
int main(void)
{
#ifdef ACM
freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif
Matrix a, ans, temp;
while(scanf("%lld%lld", &k, &mod) != EOF) {
memset(a.m, 0, sizeof(a.m));
memset(temp.m, 0, sizeof(temp.m));
for(int i = 0; i < 10; i++) {
scanf("%lld", &a.m[0][i]);
}
for(int i = 1; i < 10; i++) {
a.m[i][i-1] = 1;
}
for(int i = 0;i < 10; i++) {
temp.m[i][0] = 9 - i;
}
if(k < 10) {
printf("%lld\n", k % mod);
}
else{
ans = pow(a, k - 9);
ans = multi(ans, temp);
printf("%lld\n", ans.m[0][0]);
}
}
return 0;
}