The winner is...

本文探讨了作者对于奥巴马的支持是否源自直觉,反映了个人对于政治人物的情感倾向。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

难道我也是出于直觉,比较支持奥巴马?

来源:http://news.163.com/special/00013128/obamafoto.html


转载于:https://www.cnblogs.com/nonlyli/archive/2008/11/05/1327243.html

(c++题解,代码运行时间小于200ms,内存小于64MB,代码不能有注释)It’s time for the company’s annual gala! To reward employees for their hard work over the past year, PAT Company has decided to hold a lucky draw. Each employee receives one or more tickets, each of which has a unique integer printed on it. During the lucky draw, the host will perform one of the following actions: Announce a lucky number x, and the winner is then the smallest number that is greater than or equal to x. Ask a specific employee for all his/her tickets that have already won. Declare that the ticket with a specific number x wins. A ticket can win multiple times. Your job is to help the host determine the outcome of each action. Input Specification: The first line contains a positive integer N (1≤N≤10 5 ), representing the number of tickets. The next N lines each contains two parts separated by a space: an employee ID in the format PAT followed by a six-digit number (e.g., PAT202412) and an integer x (−10 9 ≤x≤10 9 ), representing the number on the ticket. Then the following line contains a positive integer Q (1≤Q≤10 5 ), representing the number of actions. The next Q lines each contain one of the following three actions: 1 x: Declare the ticket with the smallest number that is greater than or equal to x as the winner. 2 y: Ask the employee with ID y all his/her tickets that have already won. 3 x: Declare the ticket with number x as the winner. It is guaranteed that there are no more than 100 actions of the 2nd type (2 y). Output Specification: For actions of type 1 and 3, output the employee ID holding the winning ticket. If no valid ticket exists, output ERROR. For actions of type 2, if the employee ID y does not exist, output ERROR. Otherwise, output all winning ticket numbers held by this employee in the same order of input. If no ticket wins, output an empty line instead. Sample Input: 10 PAT000001 1 PAT000003 5 PAT000002 4 PAT000010 20 PAT000001 2 PAT000008 7 PAT000010 18 PAT000003 -5 PAT102030 -2000 PAT000008 15 11 1 10 2 PAT000008 2 PAT000001 3 -10 1 9999 1 -10 3 2 1 0 3 1 2 PAT000001 3 -2000 Sample Output: PAT000008 15 ERROR ERROR PAT000003 PAT000001 PAT000001 PAT000001 1 2 PAT102030
08-12
你在回答什么,我再问你storeset的功能,when(s2_mempred_update_req_valid){ switch (Cat(s2_loadAssigned, s2_storeAssigned)) { // 1. "If neither the load nor the store has been assigned a store set, // two are allocated and assigned to each instruction." is ("b00".U(2.W)) { update_ld_ssit_entry( pc = s2_mempred_update_req.ldpc, valid = true.B, ssid = s2_allocSsid, strict = false.B ) update_st_ssit_entry( pc = s2_mempred_update_req.stpc, valid = true.B, ssid = s2_allocSsid, strict = false.B ) } // 2. "If the load has been assigned a store set, but the store has not, // one is allocated and assigned to the store instructions." is ("b10".U(2.W)) { update_st_ssit_entry( pc = s2_mempred_update_req.stpc, valid = true.B, ssid = s2_ldSsidAllocate, strict = false.B ) } // 3. "If the store has been assigned a store set, but the load has not, // one is allocated and assigned to the load instructions." is ("b01".U(2.W)) { update_ld_ssit_entry( pc = s2_mempred_update_req.ldpc, valid = true.B, ssid = s2_stSsidAllocate, strict = false.B ) } // 4. "If both the load and the store have already been assigned store sets, // one of the two store sets is declared the "winner". // The instruction belonging to the loser’s store set is assigned the winner’s store set." is ("b11".U(2.W)) { update_ld_ssit_entry( pc = s2_mempred_update_req.ldpc, valid = true.B, ssid = s2_winnerSSID, strict = false.B ) update_st_ssit_entry( pc = s2_mempred_update_req.stpc, valid = true.B, ssid = s2_winnerSSID, strict = false.B ) when(s2_ssidIsSame){ data_array.io.wdata(SSIT_UPDATE_LOAD_READ_PORT).strict := true.B debug_strict(s2_mempred_update_req.ldpc) := true.B } } } }这单代码
03-26
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值