1165: zas的二进制
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 88 Solved: 39
Description
zas最近在玩一种特别的二进制数,它没有前导零且只由1和0组成,并且没有连续的1相邻在一起,例如前几个就是1,10,100,101,1000,1001,1010,10000.
现在zas问jzr长度为n的这种二进制数有多少个呢?因为答案会很大很大很大,请注意答案要对10007取模。
Input
第一行一个t,代表接下来有多少组数据t<=100000,
接下来t行,每行一个数n<=100000
Output
每行一个答案,输出长度为n的这种特别的二进制数有多少个,答案对10007取模。
Sample Input
3111111
Sample Output
189410
HINT
Source
思路:①直接递推,预处理。
# include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # define mod 10007 using namespace std; int dp[100001][2] = {0}; int main() { int t, n; scanf("%d",&t); dp[1][1] = 1; dp[1][0] = 0; for(int i=2; i<=100000; ++i) { dp[i][1] = dp[i-1][0]; dp[i][0] = (dp[i-1][0] + dp[i-1][1])%mod; } while(t--) { scanf("%d",&n); printf("%d\n",(dp[n][0]+dp[n][1])%mod); } return 0; }
②第n为0的情况有几种呢?有dp[n-1]种(上一个状态的情况直接补个零),第n位为1的情况有几种呢?取决于第n-1位为0的种类数,即dp[n-2],所以dp[i] = dp[i-1] + dp[i-2],恰好是斐波拉契数列。
# include <iostream> # include <cstdio> # include <cstring> # include <algorithm> # define mod 10007 using namespace std; int dp[100001]= {0}; int main() { int t, n; scanf("%d",&t); dp[1] = dp[2] = 1; for(int i=3; i<=100000; ++i) dp[i] = (dp[i-1] + dp[i-2])%mod; while(t--) { scanf("%d",&n); printf("%d\n",dp[n]); } return 0; }