timus 1180. Stone Game 解题报告

Posted 2020-06-18 IVY_BUG

1.题目：

1180. Stone Game

Time limit: 1.0 second
Memory limit: 64 MB

Two Nikifors play a funny game. There is a heap of N stones in front of them. Both Nikifors in turns take some stones from the heap. One may take any number of stones with the only condition that this number is a nonnegative integer power of 2 (e.g. 1, 2, 4, 8 etc.). Nikifor who takes the last stone wins. You are to write a program that determines winner assuming each Nikifor does its best.

Input

An input contains the only positive integer number N (condition N ≤ 10^250 holds).

Output

The first line should contain 1 in the case the first Nikifor wins and 2 in case the second one does. If the first Nikifor wins the second line should contain the minimal number of stones he should take at the first move in order to guarantee his victory.

Sample

input	output
8	1 2

2.解题思路

这种博弈问题，都是从最简单的情况考虑，递推到复杂情况的。但是这道题有一些有趣的技巧~

基本的递推：

N　　win?

1　　y

2　　y

3　　n

嗯。分析到这里，思路大概是这样的：

bool suc[maxN+1];

suc[1] = suc[2] = true;

for (i = 3; i <= n; i++) {

　　for (b = 1; b < n; b *= 2) {

　　　　if (suc[i-b] == false) {

　　　　　　suc[i] = true;

　　　　　　break;

　　　　}

　　}

}

然而再回头看一下题目规模，发现n的取值范围是n <= 10^250。这是要高精度的节奏啊。

不急，回去再看一下有没有优化的方法。通过找规律，

发现：N = 3*n， suc[n] = n

　　　N = 3*n + 1 || N = 3*n + 2， suc[n] = y

这样问题转换为求n%3。

而一个十进制数N = sum(ai*10^i) = sum(ai) + sum(ai*(10^i-1))(i>=0)

而10^i-1 = 0 (mod3)

故N = sum(ai) (mod3)

3.代码：

#include <iostream>
using namespace std;
int main()
{
    int s = 0;
    char tmp;
    while (cin >> tmp) s += tmp - ‘0‘;
    if (s % 3 == 0) cout << 2;
    else cout << 1 << ‘\n‘ << s % 3;
    //return 0;
}