题目链接:

http://exam.upc.edu.cn/problem.php?id=5121


题目:

题目描述

Alice and Bob are in their class doing drills on multiplication and division. They quickly get bored and instead decide to play a game they invented.
The game starts with a target integer N≥2, and an integer M = 1. Alice and Bob take alternate turns. At each turn, the player chooses a prime divisor p of N, and multiply M by p. If the player’s move makes the value of M equal to the target N, the player wins. If M > N, the game is a tie.
Assuming that both players play optimally, who (if any) is going to win?

输入

The first line of input contains T (1≤T≤10000), the number of cases to follow. Each of the next T lines describe a case. Each case is specified by N (2≤N≤231-1) followed by the name of the player making the first turn. The name is either Alice or Bob.

输出

For each case, print the name of the winner (Alice or Bob) assuming optimal play, or tie if there is no winner.

样例输入

10
10 Alice
20 Bob
30 Alice
40 Bob
50 Alice
60 Bob
70 Alice
80 Bob
90 Alice
100 Bob

样例输出

Bob
Bob
tie
tie
Alice
tie
tie
tie
tie
Alice

题意:

  初始的时候有一个数字$m = 1$,然后给你一个数字n,每次你可以从n的所有素因子中拿出一个乘到$m$上去,如果谁乘完之后$m$刚好等于$n$,则获胜,给你先手的人,然后输出谁能赢,如果$m>n$则输出tie


思路:

  先预处理出每个数字的所有素因子,然后开始判断,如果素因子的个数大于2则一定是平局,因为两个人都没有必胜的方法。

  如果等于2,则比较一下这两个素因子出现次数,如果相差大于1,则是平局,原因同上。如果相等则后手必胜,不管每次先手选什么,后手就选和先手相对的那个,这样的话就一定可以后手必胜,相当于是一个奇异局势。不相等就先手必胜,因为两种素因子只相差一个,拿掉多的那一个,就转化成了奇异局势。

  如果只有一个素因子的话判断一下出现的奇偶次数,奇数次就先手必胜,否则后手必胜。


实现:

#include <bits/stdc++.h>
int prime[18000], a[18000], len;
std::map<int, int> cnt;
bool check[int(2e5) + 7];
int printprime(int N) {
    memset(check, false, sizeof(check));
    int tot = 0;
    for(int i=2 ; i<=N ; ++i){
        if (! check[i]) prime[tot++] = i;
        for(int j=0 ; j<tot ; j++) {
            if(i * prime[j] > N) break;
            check [i * prime[j]] = true;
            if(i % prime[j] == 0) break;
        }
    }
    return tot;
}
char s[2][107] = {"Alice", "Bob"}, name[107];

int main() {
//    freopen("in.txt", "r", stdin);
    int tot = printprime(int(2e5));
    int _, n, fis, sec;
    scanf("%d", &_);
    while (_--) {
        cnt.clear();
        len = 0;
        scanf("%d %s", &n, name);
        if (name[0] == 'A') fis = 0, sec = 1;
        else fis = 1, sec = 0;
        int tmp = n;
        for (int i = 0; i < tot && tmp != 1; i++)
            if (tmp % prime[i] == 0)
                while (tmp % prime[i] == 0) {
                    if (cnt[prime[i]] == 0) a[len++] = prime[i];
                    cnt[prime[i]]++;
                    tmp /= prime[i];
                }
        if (tmp != 1) a[len++] = tmp, cnt[tmp]++;
        if (len == 1) {
            if (cnt[a[0]] & 1) puts(s[fis]);
            else puts(s[sec]);
            continue;
        } else if (len == 2) {
            int l = cnt[a[0]], r = cnt[a[1]];
            int x = abs(r - l);
            if (x > 1) puts("tie");
            else {
                if (l == r) puts(s[sec]);
                else puts(s[fis]);
            }
        } else puts("tie");
    }
    return 0;
}
最后修改:2018 年 04 月 25 日 12 : 27 AM
谢谢老板!