So Easy!
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5642 Accepted Submission(s): 1875
Where a, b, n, m are positive integers.┌x┐is the ceil of x. For example, ┌3.14┐=4. You are to calculate Sn.
You, a top coder, say: So easy!
總結:公式有點難推。題意給的條件(a-1)2< b < a2, 很重要
#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 = 1e5 + 10;
ll _mod;
struct Matrix
{
ll m[2][2];
};
Matrix I = {
1,0,
0,1
};
Matrix multi(Matrix a, Matrix b)
{
Matrix c;
for (int i = 0; i < 2; i++) {
for (int j = 0; j < 2; j++) {
c.m[i][j] = 0;
for (int k = 0; k < 2; 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 n)
{
Matrix ans = I;
Matrix p = a;
while (n) {
if (n & 1)
ans = multi(ans, p);
p = multi(p, p);
n /= 2;
}
return ans;
}
int main()
{
#ifdef ACM
#else
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif
ll a, b, n;
while (scanf("%lld%lld%lld%lld", &a, &b, &n, &_mod) != EOF) {
Matrix M, ans, temp, num;
M.m[0][0] = 2 * a % _mod;
M.m[0][1] = ((b % _mod - a * a % _mod) + _mod) % _mod;
M.m[1][0] = 1;
M.m[1][1] = 0;
temp.m[0][0] = 2 * (a*a%_mod + b % _mod) % _mod;
temp.m[1][0] = 2 * a % _mod;
temp.m[1][1] = 0;
temp.m[0][1] = 0;
if (n == 1)
printf("%lld\n", temp.m[1][0]);
else {
ans = pow(M, n - 2);
ans = multi(ans, temp);
printf("%lld\n", (ans.m[0][0] + _mod) % _mod);
}
}
return 0;
}