Computer Transformation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 5065 Accepted Submission(s): 1850
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.
How many pairs of consequitive zeroes will appear in the sequence after n steps?
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
1 #include <stdio.h> 2 #include <string.h> 3 4 int an1[500],an2[500]; 5 char s[1010][500]; 6 7 int main() 8 { 9 int i,j,k,len,t,n,aa; 10 memset(an1,0,sizeof(an1)); 11 memset(an2,0,sizeof(an2)); 12 memset(s,0,sizeof(s)); 13 strcpy(s[1],"0"); 14 strcpy(s[2],"1"); 15 an1[0]=1; 16 len=0;t=0; 17 for(i=3;i<=1000;i++) 18 { 19 t=0; 20 for(j=0;j<=len;j++) 21 { 22 an2[j]=an1[j]*2+t; 23 if(an2[j]>9) 24 { 25 an2[j]-=10; 26 t=1; 27 } 28 else 29 t=0; 30 if(j==len&&t==1) 31 len+=1; 32 an1[j]=an2[j]; 33 } 34 if(i&1) 35 {an1[0]=an2[0]-=1;} 36 else 37 {an1[0]=an2[0]+=1;} 38 for(k=0,aa=j-1;aa>=0;aa--) 39 { 40 an1[aa]=an2[aa]; 41 s[i][k++]=an2[aa]+'0'; 42 } 43 } 44 while(scanf("%d",&n)!=EOF) 45 { 46 printf("%s ",s[n]); 47 } 48 return 0; 49 } 50 //大数，找规律
找规律，大数问题