某島

… : "…アッカリ~ン . .. . " .. .
November 7, 2010

校賽。。


城管大隊原來是沙漠君。 。。。

#include 
using namespace std ;

int   N ;
int   a1 , a2 , s1 , s2 , F[10000001] ;

void	Init()
{
		a1 = 5 ; a2 = 401 % 29 ;
		s1 = 6 ; s2 = 402 % 29 ;
		F[1] = 5 ;
		for ( int i = 2 ; i <= 10000000 ; i ++ )
		{
			a1 = ( a1 * 5   ) % 29 ;
			a2 = ( a2 * 401 ) % 29 ;
			F[i] = ( F[i-1] + s1 * a2 + s2 * a1 + a1 * a2 ) % 29 ;
			s1 = (s1 + a1) % 29 ;
			s2 = (s2 + a2) % 29 ;
		}
}

int		main()
{
		Init() ;
		while ( cin >> N && N != 0 ) cout << F[N] << "\n" ;
		return 0 ;
}

/* This Code is Submitted by xiaodao for Problem 101449 at 2010-11-07 13:55:43 */
#include 
using namespace std;
const int N = 366;
int g[N], f[N]; double a[N];
// f:.. Cana.   g:.. Usa.
double ans;
int n;
 
 
void init(){
    for (int i=1;i<=n;i++){
        scanf("%lf", &a[i]);
		// a[i] = a[i]*0.97;
    }
}
 
void solve(){
    f[0] = 100000; g[0] = 0;
    for (int i=1;i<=n;i++){
        f[i] = max(f[i-1], int(g[i-1]*a[i]*0.97));
        g[i] = max(g[i-1], int(f[i-1]/a[i]*0.97));
    }
	
}
 
 
int main(){
    cin >> n;
    while (n!=0){
        init(); solve(); ans = double(f[n]) / 100;
        printf("%.2lf\n", ans);
        //cout << f[n]/100 << endl;
        cin >> n;
    }
	
}
#include 
#include 
using namespace std;
const int N = 10;
int A[N+1];
string s;
int x;



string f(int x){
	string res;
	
	if (x < 0) {
		res = ""; x = -x;
		
		while (x!=0){
			res = char( x%10 + 48) + res;
			x /= 10;
		}
		
		
		if (res=="") res = '0';
		
		res = "-" + res;
	}
	else {	
		while (x!=0){
			res = char(x%10 + 48) + res;
			x /= 10;
		}
		
		if (res=="") res = '0';
		
		res = "+" + res;	
	}
	
	return res;
}

string g(int x){
	
	
	if (x == 0) return "";
	if (x == 1) return "x";
	
	string res = "x^"; res += char(x + 48);
	return res;
}


void encode(){
	
	s = "+" + s;
	
	int x;
	for (int i=0;i=0;i--)
		if (A[i]!=0) s = s + f(A[i]) + g(i);
	
	
	if (s[0]=='+') s.erase(0, 1);
	
	cout << s << endl; 
}


void derivate(){
	for (int i=0;i> x; cin >> s; cout << s << endl;
}

void solve(){
	memset(A, 0, sizeof(A));
	encode(); 
	
	int i;
	for (i=1;i> T;
	for (int i=1;i<=T;i++){
		cout << "POLYNOMIAL " << i << endl;
        init(); solve();
    }
	
}

Solution for Problem A :

Problem_A_____________________________________________________________
____Description:_求2005^X的所有因數的和 mod 29________________________
____Solution___:_首先,2005是由401和5這2個質數相乘的。________________
_________________我們考慮2005^X與2005^(X-1)___________________________
_________________與2005^(X-1)相比,2005^X多出來的因數有:_____________
_____________________(1)_5^i_*_401^X__________________________________
_____________________(2)_5^X_*_401^i__________________________________
_________________於是我們用s1,s2分別記錄前X-1個5的冪和401的冪的和_____
_________________於是_F[X]_=_F[X-1]_+_s1*401^X_+_s2*5^X_+_5^X*401^X___
_________________遞推求出所有的答案,打表。時間複雜度度O(N)___________

Problem_B_____________________________________________________________
____Description:_求用1*2的骨牌和3拐的骨牌覆蓋2*N的棋盤的方案個數______
____Solution___:_用0、1、2、3表示4種狀態______________________________
____________________0、□___1、□___2、■___3、■_____________________
_______________________□______■______□______■_____________________
_________________於是我們得到下面的關係矩陣___________________________
________________________|0___1___2___3________________________________
______________________--+--------------_______________________________
______________________0_|0___1___1___1________________________________
______________________1_|0___0___1___1________________________________
______________________2_|0___1___0___1________________________________
______________________3_|1___0___0___1________________________________
_________________用快速冪矩陣乘法DP即可。_____________________________