For free

好久没写搜索了，差不多要不会了。看了下滔滔的解题报告才写出来的。http://blog.youkuaiyun.com/chuck_0430/article/details/8127339深搜的核心也就是 vis 标志数组的改变，并在中间递归调用 dfs#include #include #include #define NUMBER 500000 #define LEN 10000

Easy problem and I directly attached to the code.#include #include using namespace std;#define LEN 1000int mon[LEN];int index[LEN];int main() { int n, k, tmp; int i, count; scanf("

Brute force problem and it is very easy.My code is as follows:#include #include using namespace std;#define LEN 101long long fib[LEN];void init() { fib[0] = 0; fib[1] = 1; for(in

广搜，每个节点有两个值，一个是当前的数字，另一个是余数，用余数可以减少以后的计算量，并且可进行剪枝，从而减少了常数时间的复杂度#include #include #include #include #include using namespace std;struct Node { string num; int mod; Node() {} Node(stri