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;

}