Computer Transformation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6045 Accepted Submission(s): 2208
Problem Description
A sequence consisting of one digit, the number 1 is initially written into a computer. At each successive time step, the computer simultaneously tranforms each digit 0 into the sequence 1 0 and each digit 1 into the sequence 0 1.
So, after the first time step, the sequence 0 1 is obtained; after the second, the sequence 1 0 0 1, after the third, the sequence 0 1 1 0 1 0 0 1 and so on.
So, after the first time step, the sequence 0 1 is obtained; after the second, the sequence 1 0 0 1, after the third, the sequence 0 1 1 0 1 0 0 1 and so on.
How many pairs of consequitive zeroes will appear in the sequence after n steps?
Input
Every input line contains one natural number n (0 < n ≤1000).
Output
For each input n print the number of consecutive zeroes pairs that will appear in the sequence after n steps.
Sample Input
2 3
Sample Output
1 1
Source
/*题解：
找规律，满足a[i]=a[i-1]+2*a[i-2];
由于数据范围很大，需要用大数解决。
*/
找规律，满足a[i]=a[i-1]+2*a[i-2];
由于数据范围很大，需要用大数解决。
*/
#include<cstdio>
#include<cstring>
int a[1002][1010];
void fun(int n)
{
int i,j,c,ans,d=1;
a[0][0]=0;
a[1][0]=0;
a[2][0]=1;
a[3][0]=1;
for(i=4; i<=n; i++)
{
c=0;
for(j=0; j<d; j++)
{
ans=a[i-1][j]+2*a[i-2][j]+c;
a[i][j]=ans%10;
c=ans/10;
}
while(c)
{
a[i][d++]=c%10;
c/=10;
}
}
for(i=d-1; i>=0; i--)
printf("%d",a[n][i]);
printf("\n");
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
memset(a,0,sizeof(a));
fun(n);
}
return 0;
}