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题目链接：

# A Simple Nim

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 181    Accepted Submission(s): 119

Problem Description
Two players take turns picking candies from n heaps,the player who picks the last one will win the game.On each turn they can pick any number of candies which come from the same heap(picking no candy is not allowed).To make the game more interesting,players can separate one heap into three smaller heaps(no empty heaps)instead of the picking operation.Please find out which player will win the game if each of them never make mistakes.

Input
Intput contains multiple test cases. The first line is an integer

Output
For each test case,output a line whick contains either"First player wins."or"Second player wins".

Sample Input
2
2
4 4
3
1 2 4

Sample Output
Second player wins.
First player wins.

题意：

打表找规律的题,好想知道怎么归纳法证明啊;
一直不知道怎么打表,直接笨死了;

思路：

AC代码;

/************************************************
┆  ┏┓　　　┏┓ ┆
┆┏┛┻━━━┛┻┓ ┆
┆┃　　　　　　　┃ ┆
┆┃　　　━　　　┃ ┆
┆┃　┳┛　┗┳　┃ ┆
┆┃　　　　　　　┃ ┆
┆┃　　　┻　　　┃ ┆
┆┗━┓　  　┏━┛ ┆
┆　　┃　  　┃　　┆
┆　　┃　  　┗━━━┓ ┆
┆　　┃　 AC代马 　　┣┓┆
┆　　┃　        　　┏┛┆
┆　　┗┓┓┏━┳┓┏┛ ┆
┆　　　┃┫┫　┃┫┫ ┆
┆　　　┗┻┛　┗┻┛ ┆
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

char CH; bool F=false;
for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
if(!p) { puts("0"); return; }
while(p) stk[++ tp] = p%10, p/=10;
while(tp) putchar(stk[tp--] + '0');
putchar('
');
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const int inf=1e9;
const int N=1e6+10;
const int maxn=2e3+14;
const double eps=1e-12;
/*
int f[110],sg[110];
inline void Init()
{
For(i,0,100)
{
mst(f,0);
For(j,1,i)
{
f[sg[i-j]]=1;
}
if(i>=3)
{
For(j,1,i-1)
For(k,1,i-1)
{
if(j+k<i)f[sg[j]^sg[k]^sg[i-j-k]]=1;
}
}
For(j,0,100)if(!f[j]){sg[i]=j;break;}
cout<<i<<" "<<sg[i]<<endl;
}
}
*/
int main()
{
//Init();
int t;
while(t--)
{
int n,sum=0,x;
For(i,1,n)
{
if(x%8==7)sum^=x+1;
else if(x%8==0&&x)sum^=x-1;
else sum^=x;
}
if(sum==0)cout<<"Second player wins.
";
else cout<<"First player wins.
";
}
return 0;
}


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• 原文地址：https://www.cnblogs.com/zhangchengc919/p/5738694.html