hdoj 5451 Best Solver 【求循环节 + 矩阵快速幂】
Best SolverTime Limit: 1500/1000 MS (Java/Others) Memory Limit: 65535/102400 K (Java/Others)Total Submission(s): 229 Accepted Submission(s): 98
Problem Description
The so-called best problem solver can easily solve this problem, with his/her childhood sweetheart.
It is known that y=(5+26√)1+2x . For a given integer x (0≤x<232) and a given prime number M (M≤46337) , print [y]%M . ( [y] means the integer part of y )
Input
An integer
T (1<T≤1000) , indicating there are
T test cases.
Following are T lines, each containing two integers x and M , as introduced above.
Output
The output contains exactly
T lines.
Each line contains an integer representing [y]%M .
Sample Input
Sample Output
|
比赛时卡在这道题了，醉了。
打表找出循环节loop = (M - 1) * (M + 1)，然后求解 2^n % loop，下面就是矩阵快速幂裸题了。
提交6次代码才过，CE5次。杭电要炸了。。。
找循环节解析链接：点我
裸题链接：点我
AC代码：
#include <cstdio>
#include <cstring>
#include <cmath>
#define LL long long
struct Matrix
{
LL a[3][3];
};
Matrix ori, res;
LL M;
void init()
{
memset(ori.a, 0, sizeof(ori.a));
memset(res.a, 0, sizeof(res.a));
for(int i = 0; i < 2; i++)
res.a[i][i] = 1;
ori.a[0][0] = 5, ori.a[0][1] = 2;
ori.a[1][0] = 12, ori.a[1][1] = 5;
}
Matrix muitl(Matrix x, Matrix y)
{
Matrix z;
memset(z.a, 0, sizeof(z.a));
for(int i = 0; i < 2; i++)
{
for(int k = 0; k < 2; k++)
{
if(x.a[i][k] == 0) continue;
for(int j = 0; j < 2; j++)
z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % M) % M;
}
}
return z;
}
LL F[3], ans[3];
int k = 1;
void solve(LL n)
{
while(n)
{
if(n & 1)
res = muitl(ori, res);
ori = muitl(ori, ori);
n >>= 1;
}
for(int i = 0; i < 2; i++)
{
ans[i] = 0;
for(int k = 0; k < 2; k++)
ans[i] = (ans[i] + (F[k] * res.a[k][i]) % M) % M;
}
printf("Case #%d: %lld\n", k++, (2*ans[0] - 1) % M);
}
LL pow_mod(LL a, LL p, LL n)
{
if(p == 0) return 1;
LL ans = pow_mod(a, p/2, n);
ans = ans * ans % n;
if(p % 2 == 1) ans = ans * a % n;
return ans;
}
int main()
{
int t;
LL n, N;
F[0] = 5, F[1] = 2;
scanf("%d", &t);
while(t--)
{
scanf("%lld%lld", &n, &M);
LL loop = (M+1) * (M-1);//循环节
LL N = pow_mod(2, n, loop) + 1;//快速幂 求解2^n % loop
init();//构造矩阵
solve(N-1);//矩阵快速幂
}
return 0;
}