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重新安装浏览器，或使用别的浏览器IRLU7843PBF | Infineon HEXFET 系列 Si N沟道 MOSFET IRLU7843PBF, 161 A, Vds=30 V, 3引脚 IPAK封装 | Infineon
RS 库存编号
制造商零件编号
IRLU7843PBF
Infineon 系列分离式 HEXFET(R) 功率 MOSFET 包括 N 通道设备，采用表面安装和引线封装。 形状系数可解决大多数板布局和热设计挑战问题。 在整个范围内，基准导通电阻减少了传导损耗，让设计人员可以提供最佳系统效率。
Infineon 提供庞大且全面的 MOSFET 设备组合，其中包括 CoolMOS、OptiMOS 和 StrongIRFET 系列。它们提供同类最佳性能，实现更高效率、功率密度和成本效益。需要高质量和增强型保护功能的设计获益于符合 AEC-Q101 汽车工业标准的 MOSFET。
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最大连续漏极电流161 A
最大漏源电压30 V
最大漏源电阻值3.3 mΩ
最大栅阈值电压2.3V
最小栅阈值电压1.4V
最大栅源电压-20 V、+20 V
封装类型IPAK
安装类型通孔
晶体管配置单
通道模式增强
类别功率 MOSFET
最大功率耗散140 W
每片芯片元件数目1
宽度2.39mm
典型关断延迟时间34 ns
晶体管材料Si
尺寸6.73 x 2.39 x 7.49mm
长度6.73mm
最高工作温度+175 °C
系列HEXFET
高度7.49mm
典型输入电容值@Vds4380 pF @ 15 V
典型栅极电荷@Vgs34 nC @ 4.5 V
最低工作温度-55 °C
典型接通延迟时间25 ns
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Infineon HEXFET 系列 Si N沟道 MOSFET IRLU7843PBF, 161 A, Vds=30 V, 3引脚 IPAK封装
IRLU7843PBF
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RS 库存编号
Infineon HEXFET 系列 Si N沟道 MOSFET IRLU7843PBF, 161 A, Vds=30 V, 3引脚 IPAK封装
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制造商零件编号
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欧时电子元件（上海）有限公司
Oct 4, 2017
欧时电子元件(上海）有限公司, 上海市黄浦区延安东路618号东海商业中心二期23楼 200001
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此网站 (www.) 的管理员禁止了您的访问。原因是您的访问包含了非浏览器特征(3a848b0a-ua98).
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Time Limit:
MS (/Others)
Memory Limit:
K (Java/Others)
Total Submission(s): 4975
Accepted Submission(s): 2141
Problem Description
Arthur and his sister Caroll have been playing a game called Nim for some time now. Nim is played as follows:
The starting position has a number of heaps, all containing some, not necessarily equal, number of beads.
The players take turns chosing a heap and removing a positive number of beads from it.
The first player not able to make a move, loses.
Arthur and Caroll really enjoyed playing this simple game until they recently learned an easy way to always be able to find the best move:
Xor the number of beads in the heaps in the current position (i.e. if we have 2, 4 and 7 the xor-sum will be 1 as 2 xor 4 xor 7 = 1).
If the xor-sum is 0, too bad, you will lose.
Otherwise, move such that the xor-sum becomes 0. This is always possible.
It is quite easy to convince oneself that this works. Consider these facts:
The player that takes the last bead wins.
After the winning player's last move the xor-sum will be 0.
The xor-sum will change after every move.
Which means that if you make sure that the xor-sum always is 0 when you have made your move, your opponent will never be able to win, and, thus, you will win.
Understandibly it is no fun to play a game when both players know how to play perfectly (ignorance is bliss). Fourtunately, Arthur and Caroll soon came up with a similar game, S-Nim, that seemed to solve this problem. Each player is now only allowed to remove
a number of beads in some predefined set S, e.g. if we have S =(2, 5) each player is only allowed to remove 2 or 5 beads. Now it is not always possible to make the xor-sum 0 and, thus, the strategy above is useless. Or is it?
your job is to write a program that determines if a position of S-Nim is a losing or a winning position. A position is a winning position if there is at least one move to a losing position. A position is a losing position if there are no moves to a losing position.
This means, as expected, that a position with no legal moves is a losing position.
Input consists of a number of test cases. For each test case: The first line contains a number k (0 < k ≤ 100 describing the size of S, followed by k numbers si (0 < si ≤ 10000) describing S. The second line contains a number m (0
< m ≤ 100) describing the number of positions to evaluate. The next m lines each contain a number l (0 < l ≤ 100) describing the number of heaps and l numbers hi (0 ≤ hi ≤ 10000) describing the number of beads in the heaps. The last test case is followed by
a 0 on a line of its own.
For each position: If the described position is a winning position print a 'W'.If the described position is a losing position print an 'L'. Print a newline after each test case.
Sample Input
4 2 3 7 12
5 1 2 3 4 5
4 2 3 7 12
Sample Output
这道题的题目挺长的，大致就说，给定一个s集合，每次 行走的步数属于s集合，，，共有n堆石子，问首先出手的能否必胜
话说看了好久的SG函数，，有一丁点儿领悟~~不敢误导大家~就直接上代码了
Time Limit:
MS (Java/Others)
Memory Limit:
K (Java/Others)
Total Submission(s): 4975
Accepted Submission(s): 2141
Problem Description
Arthur and his sister Caroll have been playing a game called Nim for some time now. Nim is played as follows:
The starting position has a number of heaps, all containing some, not necessarily equal, number of beads.
The players take turns chosing a heap and removing a positive number of beads from it.
The first player not able to make a move, loses.
Arthur and Caroll really enjoyed playing this simple game until they recently learned an easy way to always be able to find the best move:
Xor the number of beads in the heaps in the current position (i.e. if we have 2, 4 and 7 the xor-sum will be 1 as 2 xor 4 xor 7 = 1).
If the xor-sum is 0, too bad, you will lose.
Otherwise, move such that the xor-sum becomes 0. This is always possible.
It is quite easy to convince oneself that this works. Consider these facts:
The player that takes the last bead wins.
After the winning player&#39;s last move the xor-sum will be 0.
The xor-sum will change after every move.
Which means that if you make sure that the xor-sum always is 0 when you have made your move, your opponent will never be able to win, and, thus, you will win.
Understandibly it is no fun to play a game when both players know how to play perfectly (ignorance is bliss). Fourtunately, Arthur and Caroll soon came up with a similar game, S-Nim, that seemed to solve this problem. Each player is now only allowed to remove a number of beads in some predefined set S, e.g. if we have S =(2, 5) each player is only allowed to remove 2 or 5 beads. Now it is not always possible to make the xor-sum 0 and, thus, the strategy above is useless. Or is it?
your job is to write a program that determines if a position of S-Nim is a losing or a winning position. A position is a winning position if there is at least one move to a losing position. A position is a losing position if there are no moves to a losing position. This means, as expected, that a position with no legal moves is a losing position.
Input consists of a number of test cases. For each test case: The first line contains a number k (0 < k ≤ 100 describing the size of S, followed by k numbers si (0 < si ≤ 10000) describing S. The second line contains a number m (0 < m ≤ 100) describing the number of positions to evaluate. The next m lines each contain a number l (0 < l ≤ 100) describing the number of heaps and l numbers hi (0 ≤ hi ≤ 10000) describing the number of beads in the heaps. The last test case is followed by a 0 on a line of its own.
For each position: If the described position is a winning position print a &#39;W&#39;.If the described position is a losing position print an &#39;L&#39;. Print a newline after each test case.
Sample Input
4 2 3 7 12
5 1 2 3 4 5
4 2 3 7 12
Sample Output
#define LEN 110
#define MAX 10010
int s[LEN] , p ,sg[MAX];
void getSG(int k)
bool hash[MAX] ;
memset(sg,0,sizeof(sg)) ;
for(int i = 0 ; i < MAX ; ++i)
memset(hash,false,sizeof(hash)) ;
for(int j = 0 ; j =0)
hash[sg[i-s[j]]] =
for(int j = 0 ; j < MAX ; ++j)
if(!hash[j])
int main()
while(~scanf("%d",&k) && k)
for(int i = 0 ; i < ++i)
scanf("%d",&s[i]) ;
scanf("%d",&m) ;
while(m--)
int temp = 0 ,
scanf("%d",&l) ;
for(int i = 0 ; i < ++i)
scanf("%d",&p) ;
temp = temp^sg[p] ;
if(temp == 0)
ans += "L";
ans += "W";
cout<<ans<<
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