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To Europe! To Europe!
Description Almost everyone in the candidate states wants to `go to Europe'', although most of the people have very vague ideas about what this actually means. Anyway, immediately after the borders are open, the inhabitants will take their cars and trucks and will `go to Europe''. This can cause many troubles, as the roads will be suddenly overloaded by vehicles of various types. You are to help to solve some of these traffic jams. Assume a convoy of vehicles has lined up in front of a single lane and one-way bridge over a river. Note that since the street is single lane, no vehicle can overtake any other. The bridge can sustain a given maximum load. To control the traffic on the bridge, operators are stationed on either end of the bridge. The convoy of vehicles is to be divided into groups, such that all the vehicles in any group can cross the bridge together. When a group reaches the other side, the operator on that side of the bridge uses a telephone to inform the operator on this side that the next group can start its journey over the bridge. The weight of each vehicle is known. The sum of the weights of the vehicles in any group cannot exceed the maximum load sustainable by the bridge. Associated with each vehicle is the maximum speed with which it can travel over the bridge. The time taken by a group of vehicles is calculated as the time taken by the slowest vehicle in the group to cross the bridge. The problem is to find the minimum amount of time in which the entire convoy can cross the bridge. Input The input consists of several test cases. The first line of each test case contains three positive integers (separated by blanks): the first one represents the maximum load that the bridge can sustain b (in tonnes); the second one represents the length of the bridge l (in kms); and the third one is the number of vehicles (n) in the convoy. Each of the next n lines of input contains a pair of positive integers, wi and si (separated by blanks), where wi is the weight of the vehicle (in tonnes) and si is the maximum speed (in kmph) with which this vehicle can travel over the bridge. The weights and speeds of the vehicles are specified in the same order as the order in which the vehicles are queued up. You can assume that 1 <= n,b,l,s <= 1000 and any i in [1..n]: wi <= b. After the last vehicle, the next test case description begins. The last test case is followed by a line containing three zeros. Output The output of the program should be a single real number specifying the minimum time in minutes in which the convoy can cross the bridge. The number should be displayed with one digit after the decimal point. Sample Input 100 5 10 40 25 50 20 50 20 70 10 12 50 9 70 49 30 38 25 27 50 19 70 0 0 0 Sample Output 75.0

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64位LINUX X86关于DMA API的一些问题， 求高手不吝赐教，急急急！
pSendDmaPool = dma_pool_create("DmaPool",pci_dev->dev), (size_t)(512 * 64*1024+1024), (size_t)(64*1024), (size_t)0); pDmaBuf = (unsigned char*)dma_pool_alloc(pSendDmaPool, GFP_KERNEL, &dma_handle); 我的平台是64位X86，我执行完dma_pool_create后，再执行dma_pool_alloc，就崩溃了，然后打印信息是： ------------[ cut here ]------------ WARNING: at mm/page_alloc.c:2139 __alloc_pages_nodemask+0x7a8/0x8d0() (Not tainted) Hardware name: All Series Modules linked in: fh130drv(U) ebtable_nat ebtables ipt_MASQUERADE iptable_nat nf_nat xt_CHECKSUM iptable_mangle bridge autofs4 8021q garp stp llc cpufreq_ondemand acpi_cpufreq freq_table mperf ipt_REJECT nf_conntrack_ipv4 nf_defrag_ipv4 iptable_filter ip_tables ip6t_REJECT nf_conntrack_ipv6 nf_defrag_ipv6 xt_state nf_conntrack ip6table_filter ip6_tables ipv6 vhost_net macvtap macvlan tun kvm uinput ppdev iTCO_wdt iTCO_vendor_support microcode parport_pc parport r8169 mii sg serio_raw lpc_ich mfd_core i2c_i801 shpchp snd_hda_codec_realtek snd_hda_codec_hdmi snd_hda_intel snd_hda_codec snd_hwdep snd_seq snd_seq_device snd_pcm snd_timer snd soundcore snd_page_alloc ext4 jbd2 mbcache sd_mod crc_t10dif ahci xhci_hcd i915 drm_kms_helper drm i2c_algo_bit i2c_core video output dm_mirror dm_region_hash dm_log dm_mod [last unloaded: scsi_wait_scan] Pid: 3725, comm: fh130demo Not tainted 2.6.32-504.el6.x86_64 #1 Call Trace: [<ffffffff81074df7>] ? warn_slowpath_common+0x87/0xc0 [<ffffffff81074e4a>] ? warn_slowpath_null+0x1a/0x20 [<ffffffff811343e8>] ? __alloc_pages_nodemask+0x7a8/0x8d0 [<ffffffff815300da>] ? atomic_notifier_call_chain+0x1a/0x20 [<ffffffff813429ae>] ? notify_update+0x2e/0x30 [<ffffffff813443b0>] ? vt_console_print+0x260/0x330 [<ffffffff810a4c2f>] ? up+0x2f/0x50 [<ffffffff81074ffa>] ? _call_console_drivers+0x4a/0x80 [<ffffffff81012fe6>] ? dma_generic_alloc_coherent+0xa6/0x160 [<ffffffff8100bb8e>] ? apic_timer_interrupt+0xe/0x20 [<ffffffff81041fd1>] ? x86_swiotlb_alloc_coherent+0x31/0x70 [<ffffffff81164f5c>] ? dma_pool_alloc+0xfc/0x2a0 [<ffffffff815293c5>] ? printk+0x41/0x44 [<ffffffffa0546b36>] ? hbaDevAllocDmaBuffers+0x2f6/0x8f0 [fh130drv] [<ffffffffa0547298>] ? hbaDevInitialize+0x168/0x1b0 [fh130drv] [<ffffffffa053fd44>] ? Hba_FC_ASM_LoadConfig+0x54/0xa0 [fh130drv] [<ffffffffa053e532>] ? hba_dev_ioctl+0x1e2/0xbd0 [fh130drv] [<ffffffff81234be1>] ? avc_has_perm+0x71/0x90 [<ffffffff8133a483>] ? pty_write+0x73/0x80 [<ffffffff812369d4>] ? inode_has_perm+0x54/0xa0 [<ffffffff8109eefc>] ? remove_wait_queue+0x3c/0x50 [<ffffffff8105bd23>] ? __wake_up+0x53/0x70 [<ffffffff811a3752>] ? vfs_ioctl+0x22/0xa0 [<ffffffff811a38f4>] ? do_vfs_ioctl+0x84/0x580 [<ffffffff811a3e71>] ? sys_ioctl+0x81/0xa0 [<ffffffff810e5a7e>] ? __audit_syscall_exit+0x25e/0x290 [<ffffffff8100b072>] ? system_call_fastpath+0x16/0x1b ---[ end trace 03df1cd6b697aafa ]--- 求高手解答这个是怎么回事？ 然后我想问一下，我可以分配 512 * 64*1024+1024这么大的DMA内存吗？并且对齐方式是64*1024？ 然后我alloc的时候，每一块是多大呢？ 谢谢了，没有金币了，不好意思！

HashMap容器从字面的理解就是，基于Hash算法构造的Map容器。从数据结构的知识体系来说，HashMap容器是散列表在Java中的具体表达（并非线性表结构）。具体来说就是，利用K-V键值对中键对象的某个属性（默认使用该对象的“内存起始位置”这一属性）作为计算依据进行哈希计算（调用hashCode方法），然后再以计算后的返回值为依据，将当前K-V键值对在符合HashMap容器构造原则的基础上，放置到HashMap容器的某个位置上，且这个位置和之前添加的K-V键值对的存储位置完全独立，不一定构成连续的存储
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