【矩阵快速幂】 HDOJ 5015 233 Matrix

构造矩阵,进行矩阵快速幂即可。。。

#include   
#include   
#include   
#include   
#include   
#include   
#include   
#include   
#include   
#include 
#include 
#include 
#include 
#define maxn 55
#define maxm 200005
#define eps 1e-10
#define mod 10000007
#define INF 1e9
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid  
#define rson o<<1 | 1, mid+1, R  
typedef long long LL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}
LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}
// head

int num[maxn], ans[maxn];
int mat[maxn][maxn], mid[maxn][maxn], res[maxn][maxn];
int n, m, t, tt;
void init(void)
{
	memset(mat, 0, sizeof mat);
	memset(res, 0, sizeof res);
	memset(ans, 0, sizeof ans);
}
void read(void)
{
	t = n + 2;
	num[1] = 1, num[2] = 233;
	for(int i = 3; i <= t; i++) scanf("%d", &num[i]);
}
void calculate(void)
{
	while(m) {
		if(m % 2) {
			for(int i = 1; i <= t; i++)
				for(int j = 1; j <= t; j++) {
					LL tmp = 0;
					for(int k = 1; k <= t; k++)
						tmp = (tmp + (LL)res[i][k] * mat[k][j]) % mod;
					mid[i][j] = tmp;
				}
			for(int i = 1; i <= t; i++)
				for(int j = 1; j <= t; j++)
					res[i][j] = mid[i][j];
		}
		for(int i = 1; i <= t; i++)
			for(int j = 1; j <= t; j++) {
				LL tmp = 0;
				for(int k = 1; k <= t; k++)
					tmp = (tmp + (LL)mat[i][k] * mat[k][j]) % mod;
				mid[i][j] = tmp;
			}
		for(int i = 1; i <= t; i++)
			for(int j = 1; j <= t; j++)
				mat[i][j] = mid[i][j];
		m /= 2;
	}
}
void work(void)
{
	mat[1][1] = 1, mat[1][2] = 3, mat[2][2] = 10;
	for(int i = 3; i <= t; i++)
		for(int j = 2; j <= i; j++)
			mat[j][i] = 1;
	for(int i = 1; i <= t; i++) res[i][i] = 1;
	calculate();
	for(int i = 1; i <= t; i++)
		for(int j = 1; j <= t; j++)
			ans[i] = ((LL)ans[i] + (LL)num[j] * res[j][i]) % mod;
	printf("%d\n", ans[t]);
}
int main(void)
{
	while(scanf("%d%d", &n, &m)!=EOF) {
		init();
		read();
		work();
	}
	return 0;
}


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