hdu 1041 Computer Transformation
Computer Transformation
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8419 Accepted Submission(s):
3165
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
大整数的题目 已经不想说什么了
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <stdio.h> 2 #include <string.h> 3 #include <iostream> 4 using namespace std; 5 int ans[1000][100]; 6 int main() 7 { 8 int n; 9 memset(ans,0,sizeof(ans)); 10 for(int i=2; i<=1000; i++) 11 { 12 for(int j=0; j<100; j++) 13 ans[i][j]=ans[i-1][j]; 14 int t=0; 15 if(i&1) 16 { 17 for(int j=0; j<100; j++) 18 { 19 t=t+ans[i][j]*2; 20 ans[i][j]=t%1000000; 21 t/=1000000; 22 } 23 ans[i][0]-=1; 24 } 25 else 26 { 27 for(int j=0; j<100; j++) 28 { 29 t=t+ans[i][j]*2; 30 ans[i][j]=t%1000000; 31 t/=1000000; 32 } 33 ans[i][0]+=1; 34 } 35 } 36 while(scanf("%d",&n)!=EOF) 37 { 38 int k=99; 39 if(n==1) 40 { 41 puts("0"); 42 continue; 43 } 44 while(ans[n][k]==0) k--; 45 printf("%d",ans[n][k]); 46 k--; 47 for(int j=k; j>=0; j--) 48 printf("%06d",ans[n][j]); 49 putchar(10); 50 } 51 return 0; 52 }
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