1sting
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4983 Accepted Submission(s): 1847
Problem Description
You will be given a string which only contains ‘1’; You can merge two adjacent ‘1’ to be ‘2’, or leave the ‘1’ there. Surly, you may get many different results. For example, given 1111 , you can get 1111, 121, 112,211,22. Now, your
work is to find the total number of result you can get.
Input
The first line is a number n refers to the number of test cases. Then n lines follows, each line has a string made up of ‘1’ . The maximum length of the sequence is 200.
Output
The output contain n lines, each line output the number of result you can get .
Sample Input
3 1 11 11111
Sample Output
1 2 8
Author
z.jt
Source
这题是斐波那契的运用,但是fib200会超掉储存,所以用大数的方式储存。
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <queue> #include <stack> #include <iostream> #define MAX_N 205 #define MAX(a, b) (a > b)? a: b #define MIN(a, b) (a < b)? a: b using namespace std; int fib[MAX_N][MAX_N]; void init() { fib[1][0] = 1, fib[2][0] = 2; for (int i = 3; i < MAX_N; i++) { int q = 0, p = 0; for (int j = 0; j < MAX_N; j++) { q = fib[i - 1][j] + fib[i - 2][j] + p; fib[i][j] = q%10; p = q / 10; } } } int main() { int t; init(); char s[MAX_N]; scanf("%d", &t); while (t--) { scanf("%s", &s); int len = strlen(s); int i; for (i = 200; i >= 0 ; i--) { if (fib[len][i] != 0) { break; } } for (int j = i; j >= 0 ; j--) { printf("%d", fib[len][j]); } printf(" "); } return 0; }