HDOJ 5088 Revenge of Nim II 位运算
2014-11-02 11:14
411 查看
位运算。。。。
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 229 Accepted Submission(s): 79
Problem Description
Nim is a mathematical game of strategy in which two players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come from the same heap.
---Wikipedia
Today, Nim takes revenge on you, again. As you know, the rule of Nim game is rather unfair, only the nim-sum (⊕) of the sizes of the heaps is zero will the first player lose. To ensure the fairness of the game, the second player has a chance to move some (can
be zero) heaps before the game starts, but he has to move one heap entirely, i.e. not partially. Of course, he can’t move all heaps out, at least one heap should be left for playing. Will the second player have the chance to win this time?
Input
The first line contains a single integer T, indicating the number of test cases.
Each test case begins with an integer N, indicating the number of heaps. Then N integer Ai follows, indicating the number of each heap.
[Technical Specification]
1. 1 <= T <= 100
2. 1 <= N <= 1 000
3. 1 <= Ai <= 1 000 000 000 000
Output
For each test case, output “Yes” if the second player can win by moving some (can be zero) heaps out, otherwise “No”.
Sample Input
3
1
2
3
2 2 2
5
1 2 3 4 5
Sample Output
No
Yes
Yes
HintFor the third test case, the second player can move heaps with 4 and 5 objects out, so the nim-sum of the sizes of the left heaps is 1⊕2⊕3 = 0.
Source
BestCoder Round #16
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long int LL;
int n;
LL c[60];
LL a[1100];
int main()
{
int T_T;
scanf("%d",&T_T);
while(T_T--)
{
memset(c,0,sizeof(c));
scanf("%d",&n);
for(int i=1;i<=n;i++)
cin>>a[i];
bool flag=false;
for(int i=1;i<=n;i++)
{
for(int j=60;j>=0;j--)
{
if(c[j]==0&&(a[i]&(1LL<<j)))
{
c[j]=a[i];
break;
}
else if(a[i]&(1LL<<j))
{
a[i]^=c[j];
if(a[i]==0) flag=true;
}
}
}
if(flag==true) puts("Yes");
else puts("No");
}
return 0;
}
Revenge of Nim II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 229 Accepted Submission(s): 79
Problem Description
Nim is a mathematical game of strategy in which two players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come from the same heap.
---Wikipedia
Today, Nim takes revenge on you, again. As you know, the rule of Nim game is rather unfair, only the nim-sum (⊕) of the sizes of the heaps is zero will the first player lose. To ensure the fairness of the game, the second player has a chance to move some (can
be zero) heaps before the game starts, but he has to move one heap entirely, i.e. not partially. Of course, he can’t move all heaps out, at least one heap should be left for playing. Will the second player have the chance to win this time?
Input
The first line contains a single integer T, indicating the number of test cases.
Each test case begins with an integer N, indicating the number of heaps. Then N integer Ai follows, indicating the number of each heap.
[Technical Specification]
1. 1 <= T <= 100
2. 1 <= N <= 1 000
3. 1 <= Ai <= 1 000 000 000 000
Output
For each test case, output “Yes” if the second player can win by moving some (can be zero) heaps out, otherwise “No”.
Sample Input
3
1
2
3
2 2 2
5
1 2 3 4 5
Sample Output
No
Yes
Yes
HintFor the third test case, the second player can move heaps with 4 and 5 objects out, so the nim-sum of the sizes of the left heaps is 1⊕2⊕3 = 0.
Source
BestCoder Round #16
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long int LL;
int n;
LL c[60];
LL a[1100];
int main()
{
int T_T;
scanf("%d",&T_T);
while(T_T--)
{
memset(c,0,sizeof(c));
scanf("%d",&n);
for(int i=1;i<=n;i++)
cin>>a[i];
bool flag=false;
for(int i=1;i<=n;i++)
{
for(int j=60;j>=0;j--)
{
if(c[j]==0&&(a[i]&(1LL<<j)))
{
c[j]=a[i];
break;
}
else if(a[i]&(1LL<<j))
{
a[i]^=c[j];
if(a[i]==0) flag=true;
}
}
}
if(flag==true) puts("Yes");
else puts("No");
}
return 0;
}
相关文章推荐
- HDOJ 5088 Revenge of Nim II 位运算
- HDU - 5088 Revenge of Nim II
- 【HDU 5088】Revenge of Nim II
- hdu 5088 Revenge of Nim II(高斯消元)
- hdu 5088 Revenge of Nim II(BestCoder Round #16)
- 【HDU】 5088 Revenge of Nim II 高斯消元
- HDU 5088 Revenge of Nim II(尼姆博弈,高斯消元)
- 【HDU】5088 Revenge of Nim II 高斯消元
- HDU 5088 Revenge of Nim II 高斯消元(异或,矩阵秩)
- hdu 5088 Revenge of Nim II(高斯消元)
- HDU 5088 Revenge of Nim II (高斯消元??)
- HDOJ 4994 Revenge of Nim
- HDOJ-5087-Revenge of LIS II
- HDOJ 5021 Revenge of kNN II
- HDOJ 5087 Revenge of LIS II DP
- HDOJ 5087 Revenge of LIS II DP
- 【BestCoder】 HDOJ 5021 Revenge of kNN II
- HDOJ 题目4099 Revenge of Fibonacci(大数, 字典树)
- BC D Revenge of kNN II hdu 5021
- 【HDU 5021】 Revenge of kNN II