题意:给两个数 n,m,让你把它们分成 全是1,每次操作只能分成几份相等的,求哪一个分的次数最多。
析:很明显,每次都除以最小的约数是最优的。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> //#include <tr1/unordered_map> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; //using namespace std :: tr1; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e4 + 5; const LL mod = 10000000000007; const int N = 1e6 + 5; const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1}; const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); } int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int a[3205]; vector<int> prime; int main(){ freopen("funny.in", "r", stdin); freopen("funny.out", "w", stdout); m = sqrt(3200+0.5); memset(a, 0, sizeof a); for(int i = 2; i <= m; ++i) if(!a[i]){ for(int j = i*i; j <= 3200; j += i) a[j] = 1; } for(int i = 2; i <= 3200; ++i) if(!a[i]) prime.push_back(i); while(scanf("%d %d", &n, &m) == 2 && m+n){ if(m == n){ printf("%d %d - Harry ", n, m); continue; } printf("%d %d - ", n, m); int ans1 = 0, ans2 = 0; for(int i = 0; i < prime.size(); ++i){ while(n % prime[i] == 0) n /= prime[i], ++ans1; while(m % prime[i] == 0) m /= prime[i], ++ans2; if(1 == n && 1 == m) break; } if(n != 1) ++ans1; if(m != 1) ++ans2; if(ans1 <= ans2) puts("Harry"); else puts("Vera"); } return 0; }