Gym 100512F Funny Game (博弈+数论)

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", 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;
}