Sicily 1305 Who’s Winner?

最新推荐文章于 2016-03-16 13:28:34 发布

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本文介绍了一个简单的二人轮流操作的游戏，通过分析游戏规则和胜利条件，给出了一种有效的判断赢家的方法。通过对不同数值范围内的游戏状态进行归纳总结，实现了对游戏胜负结果的快速判断。

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Nic and Susan play the game of multiplication by multiplying an integer p by one of the numbers 2 to 9. Nic always starts with p=1, does his multiplication. Then Susan multiplies the number, then Nic and so on. Before a game starts, they draw an integer 1≤n<4,294,967,295 and the winner is who first reaches p≥n.

Input

Each line of input contains one integer number n.

Output

For each line of input output one line either

Nic wins.

Susan wins.

Assume that both of them play perfectly.

Sample Input

162
17
34012226

Sample Output

Nic wins.
Susan wins.
Nic wins.

Problem Source

ZSUACM Team Member

Solution

这题意有点表述不清，简单来说就是p=1，然后两人先后从2-9中取数字乘上p，Nic先手，谁的操作使得p>=n谁就赢。给出n，问谁赢。

从极限触发列举情况：

n = [1.9] Nic Win

n = [10,18] SuSan Win

n = [19,162] Nic Win

在列举的时候可以知道，Nic有希望的时候就是9*2*9，会取9，而Susan会选2来阻止，同理，当SuSan有希望的时候是2*9*2*9，Nic会取2来阻止。

归纳后模拟就ok啦~

#include <cstdio>

int main()
{
  long long n;

  while (scanf("%lld", &n) == 1)
  {
    long long p = 1;

    while (p)//模拟
    {
      p *= 9;
      if (p >= n)
      {
        printf("Nic wins.\n");
        break;
      }
      p *= 2;
      if (p >= n)
      {
        printf("Susan wins.\n");
        break;
      }
    }
  }

  return 0;
}