hdu 2509 Be the Winner
2017-12-02 09:21
337 查看
点击打开链接
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4117 Accepted Submission(s): 2282
Problem Description
Let's consider m apples divided into n groups. Each group contains no more than 100 apples, arranged in a line. You can take any number of consecutive apples at one time.
For example "@@@" can be turned into "@@" or "@" or "@ @"(two piles). two people get apples one after another and the one who takes the last is
the loser. Fra wants to know in which situations he can win by playing strategies (that is, no matter what action the rival takes, fra will win).
Input
You will be given several cases. Each test case begins with a single number n (1 <= n <= 100), followed by a line with n numbers, the number of apples in each pile. There is a blank line between cases.
Output
If a winning strategies can be found, print a single line with "Yes", otherwise print "No".
Sample Input
2
2 2
1
3
Sample Output
No
Yes
anti-nim游戏:谁拿走最后一个谁输。
结论:先手胜当且仅当 (1).所有堆石子数都为1且游戏的SG值为0。(2).存在某堆石子数大于1且SG的值不为0
证明:
(1).若所有堆石子数都为1且SG值为0,则共有偶数堆石子,所以先手胜。
(2).a.只有一堆石子数大于1时,我们可以对该堆石子进行操作,使操作后石子堆数为奇数且所有堆得石子数均为1
b.有超过一堆石子数大于1时,先手将SG值变为0即可,且总存在某堆石子数大于1
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int n;
int s[110];
while(cin>>n)
{
int t,flag=0;
for(int i=0;i<n;i++)
{
cin>>s[i];
if(i==0)
t=s[0];
else
t^=s[i];
if(s[i]>1)
flag=1;
}
if(!flag)
{
if(n&1)
printf("No\n");
else
printf("Yes\n");
continue;
}
if(t==0)
printf("No\n");
else
printf("Yes\n");
}
return 0;
}
Be the Winner
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4117 Accepted Submission(s): 2282
Problem Description
Let's consider m apples divided into n groups. Each group contains no more than 100 apples, arranged in a line. You can take any number of consecutive apples at one time.
For example "@@@" can be turned into "@@" or "@" or "@ @"(two piles). two people get apples one after another and the one who takes the last is
the loser. Fra wants to know in which situations he can win by playing strategies (that is, no matter what action the rival takes, fra will win).
Input
You will be given several cases. Each test case begins with a single number n (1 <= n <= 100), followed by a line with n numbers, the number of apples in each pile. There is a blank line between cases.
Output
If a winning strategies can be found, print a single line with "Yes", otherwise print "No".
Sample Input
2
2 2
1
3
Sample Output
No
Yes
anti-nim游戏:谁拿走最后一个谁输。
结论:先手胜当且仅当 (1).所有堆石子数都为1且游戏的SG值为0。(2).存在某堆石子数大于1且SG的值不为0
证明:
(1).若所有堆石子数都为1且SG值为0,则共有偶数堆石子,所以先手胜。
(2).a.只有一堆石子数大于1时,我们可以对该堆石子进行操作,使操作后石子堆数为奇数且所有堆得石子数均为1
b.有超过一堆石子数大于1时,先手将SG值变为0即可,且总存在某堆石子数大于1
#include<iostream>
#include<cstdio>
using namespace std;
int main()
{
int n;
int s[110];
while(cin>>n)
{
int t,flag=0;
for(int i=0;i<n;i++)
{
cin>>s[i];
if(i==0)
t=s[0];
else
t^=s[i];
if(s[i]>1)
flag=1;
}
if(!flag)
{
if(n&1)
printf("No\n");
else
printf("Yes\n");
continue;
}
if(t==0)
printf("No\n");
else
printf("Yes\n");
}
return 0;
}
相关文章推荐
- HDU 2509 Be The Winner ( Nim 博弈 )
- hdu 2509 Be the Winner
- [博弈]HDU 2509 Be the Winner
- HDU 2509 Be the Winner
- HDU 2509 Be the Winner
- hdu 2509 Be the Winner(博弈Nim)
- HDU 2509 Be the Winner (尼姆博弈)
- HDU(2509) Be the Winner /HDU(1907) John (Nim 博弈)
- HDU 2509 Be the Winner
- hdu 1907——John & hdu 2509——Be the Winner
- HDU 2509 Be the Winner
- HDU-2509-Be the Winner,博弈题~~水过~~
- HDU 2509 Be the Winner && HDU 1907 John (Nim博弈变形)
- HDU-2509-Be the Winner【nimm】
- HDU 2509 Be the Winner(尼姆模板与hdu 1907类似的题目)
- HDU 2509 Be the Winner (Nim博弈)
- HDU 2509 Be the Winner (Anti-Nim)
- hdu 2509 Be the Winner(博弈)
- hdu 1907 John&& hdu 2509 Be the Winner(基础nim博弈)
- hdu 2509 Be the Winner (Anti-SG游戏+Multi-SG游戏)