[恢]hdu 1250

2011-12-26 00:33:25

地址:http://acm.hdu.edu.cn/showproblem.php?pid=1250

题意:F[1] = F[2] = F[3] = F[4] = 1。F[n] = F[n-1]+F[n-2]+F[n-3]+F[n-4]。输入n,求F[n]。大数运算。

代码:

# include <stdio.h>


int a[6][2100] ;
int buff[2100] ;


void add(int x[], int y[])
{
int i, *p, *q, cc = 0 ;
if (x[0] < y[0]) p = x, q = y ;
else p = y, q = x ;

for (i = 1 ; i <= q[0] ; i++)
{
if (i <= p[0]) buff[i] = p[i] ;
else buff[i] = 0 ;
buff[i] += q[i] + cc ;
cc = buff[i] / 10 ;
buff[i] %= 10 ;
}
if (cc != 0) buff[i++] = cc ;
buff[0] = i-1 ;
for (i = 0 ; i <= buff[0] ; i++)
x[i] = buff[i] ;
}


void cpy(int a[], int b[])
{
int i ;
for (i = 0 ; i <= b[0] ; i++)
a[i] = b[i] ;
}


void output (int a[])
{
int i ;
for (i = a[0] ; i>= 1 ; i--)
printf ("%d", a[i]) ;
printf ("\n") ;
}


int main ()
{
int n, i, j ;
while (~scanf ("%d", &n))
{
if (n <= 4){
puts ("1") ;
continue ;
}
for (i = 0 ; i < 4 ; i++)
a[i][0] = a[i][1] = 1 ;
for (i = 0 ; i < n -4 ; i++)
{
a[4][0] = 1, a[4][1] = 0 ;
for (j = 0 ; j < 4 ; j++)
add(a[4], a[j]) ;
for (j = 1 ; j < 5 ; j++)
cpy (a[j-1], a[j]) ;
}
output (a[4]) ;
}
return 0 ;
}



原文地址:https://www.cnblogs.com/lzsz1212/p/2315355.html