2018计算机考研408真题和答案.pdf

20181402801S1S2F()1S1ab2S2op3b op a4S1S15,8,3,22S2*,-,+3F()S1A.-15B.15C.-20D.202.QSQ1,2,3,4,5,61S3A.1,2,5,6,4,3B.2,3,4,5,6,1C.3,4,5,6,1,2D.6,5,4,3,2,13.1212Mmi,jCNm6,6NA.50B.51C.55D.664.T2TkTA.2k-1B.2kC.k2D.2k-15.a,b,c,d,e,f6,3,8,2,10,4A.00,1011,01,1010,11,100B.00,100,110,000,0010,01C.10,1011,11,0011,00,010D.0011,10,11,0010,01,0006.A.x1x2x5B.x1x4x5C.x3x5x4D.x4x3x57.A.1,5,2,3,6,4B.5,1,2,6,3,4C.5,1,2,3,6,4D.5,2,1,6,3,48.53BA.15B.31C.62D.2429.7HTH(k)=k%722,43,15HTA.1.5B.1.6C.2D.310.8,3,9,11,2,1,4,7,5,10,61,3,7,5,2,6,4,9,11,10,81,2,6,4,3,7,5,8,11,10,9A.3,1B.3,2C.5,2D.5,311.6,1,5,9,8,4,7)12.A.B.C.D.13.intxyFFFF FFDFH0000 0041Hxyx-yA.x=-65,y=41,x-yB.x=-33,y=65,x-yFFFF FF9DHC.x=-33,y=65,x-yFFFF FF9EHD.x=-65,y=41,x-yFFFF FF96H14.IEEE 754A.1.02-126B.1.02-127C.1.02-128D.1.02-14915.32(Little Endian)-A.C7 45 FC C0 FF FF FFB.C7 45 FC 0C FF FF FFC.C7 45 FC FF FF FF C0D.C7 45 FC FF FF FF 0C16.x1101 1000 x11A.1110 11001110 1100B.0110 11001110 1100C.1110 11000110 1100D.0110 11000110 110017.DRAMrc2K1DRAMrcA.20481B.6432C.3264D.1204818.doubleA2000HA0sizeof(double)12100HA.25B.32C.64D.10019.R1-R2/CFOFR1=FFFF FFFFHR2=FFFF FFF0HCFOF A.CF=0,OF=0B.CF=1,OF=0C.CF=0,0F=1D.CF=1,OF=120.5AE80ps50ps50ps70ps50ps20psCPUA.60 ps B.70 psC.80 psD.100 ps21./A.B.C.D.22.I/OA.B.CPUC.CPUD.CPU23.CPUA.B.C.D.24.1sT3P1P2P3CPUCPUP130s12s10P215s24s30P318s36s20CPUT A.54sB.73sC.74sD.75s25.thread1thread20 x thread1thread2x1thread1thread2mov R1,x /xinc R1 /R1mov x,R1 /R1mov R2,x /xinc R2 /R2mov x,R2 /R2x2A.1B.2C.3D.426.4P1P2P3431P1P2P3210A.B.C.P3P1P2D.P3P2P127.P .P.P.CPUA.B.C.D.28.xx.wait()A.xB.xC.xD.x29.CPU.A.B.C.D.30.A.FCFSB.SSTFC.SCAND.CSCAN31.A.B.C.D.32.A.PetersonB.swapC.D.TestAndSet33.TCP/IPA.FTPB.DNSC.SMTPD.HTTP34.A.B.C.D.35.IEEE 802.11MACCSMA/CAA.B.C.MACD.RTSCTS36.-3 kbps200ms40%A.240B.400C.480D.80037.RS1S2RH1H2IPMACH1H21IPPH1PMACH2PMACA.00-a1-b2-c3-d4-6200-1a-2b-3c-4d-52B.00-a1-b2-c3-d4-6200-a1-b2-c3-d4-61C.00-1a-2b-3c-4d-5100-1a-2b-3c-4d-52D.00-1a-2b-3c-4d-5100-a1-b2-c3-d4-6138.435.230.32.0/2135.230.40.0/2135.230.48.0/2135.230.56.0/214A.35.230.0.0/19B.35.230.0.0/20C.35.230.32.0/19D.35.230.32.0/2039.UDP(demultiplexing)A.B.C.D.40.SMTPA.JPEGB.MPEGC.EXED.ASCII41477041.(13),-5,3,2,311,2,3412CC+342.(12)BJCSXAQDJNNJTLWH8421242131H1TLH2BJH1H2TTL=5IPH2IP43.(8)500MHzCPI4AB2MB/s40MB/sI/O321AI/O/10ACPUA/CPU2I/O400BI/O3BDMADMA1000BCPUDMA500CPUB/CPU44.(15)CPU4444441?2TLBTLBSRAMDRAM3CacheCacheLRUWrite BackCacheDataTagCacheCache4CPU0008 C040HCacheCPU0007 C260HCache45.(8)4416682PDBRPDBRPDBR3CLOCK46.(7)4KB64B11814B121M1M=220512M5600B3F16KBF240KBF1F247.(7)47IP192.168.1.0/24IPMTU=1500BMTU=800B1IP2192.168.1.1192.168.1.2081500BIPIP20BF1IPIP20181B2C3A4A5A6C7D8B9C10D11A12D13C14A15A16B17C18B19A20D21B22C23C24D25B26A27C28D29D30A31D32C33B34C35D36D37D38C39B40D411BnA1nB01Bn-1nB0An1n+1An1nn+1A0n1n1nA0nA0A0Ai=nBAi-1=1ABBi=0ii+1A1nBi0i+1i=ni+1n+1An+1int findMissMin(int A,int n)int i,*B;/B=(int*)malloc(sizeof(int)*n);/memset(B,0,sizeof(int)*n);/0 for(i=0;i0&Ai=n)/Ai1nB BAi-1=1;for(i=0;in;i+)/B if(Bi=0)break;return i+1;/3ABO(1)O(n)BnO(n)421primkruskal162Primkruskal3TTL=5IP51TLBJTTL=5IPH1H2H2IP2TLBJH2IP431A324B/2MB=2us2us1s/2us=5105CPUA/5105104=2107CPU2107/500M=4%24001/500M=0.8usB32B324B/40MB=0.1usBI/O3DMACPUDMABDMA40MB/1000B=40000CPUB/40000500=2107CPU2107/500M=4%44116+12=2820+3+5=282TLBTLBTLBTLBSRAM3Cache22LRU1LRU1128Tag2035Cache23=8225=32BCache8220+1+1+1+328=4464=558CacheCache4TLB008CHTLB0040H13040H0040040H0040040H2000400H500000B3010B200400HCache2Cache00400H0Cache00400HCache12120010 0110 0000B01100000B011BCache345132201210100180 6008H2PDBRPage-Directory Base RegisterPDBRPDBRPDBRPDBRPDBR3CLOCK4614KB4B4KB/4B=10248+11024+110242+1102434KB(8+11024+110242+110243)4KB=32KB+4MB+4GB+4TB21M4KB/64B=64M5600B2512M512M/2=256M64M5600B3F16KB4KB8=32KBF1F240KB4KB840KB4KB8+4KB1024F247110192.168.1.0/24IPIP248101192.168.1.1281192.168.1.127IP27-2=12601208-129+1=801IP(192.168.1.254)126-80-1=452MTU8BMTU=800BIP20BIP(800-20)/8 8=776(1500-20)/776=2102776/8=97