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2013计算机考研408真题和答案.pdf
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2013 计算机 考研 408 答案
2013 1402801.mnm+nA.()O nB.()O m nC.(min(,)Om nD.(max(,)Om n2.1,2,3,n123,np ppp23p3pA.3nB.2nC.1nD.3.1234567TT0A.0B.1C.2D.34.T6234567TA.27B.46C.54D.565.XXYXA.XB.YC.XYD.Y6.T1vT2vT2T3T1T3I.vT1T1T3II.vT1T1T3III.vT1T1T3IV.vT1T1T3A.IIIIB.IIVC.IIIIID.IIIV7.A0101001101001000AA.1212B.2211C.3423D.44228.A.hcabdegfB.eafgbhcdC.dbcahefgD.abcdhefg9.AOE8A.ce B.de C.fd D.fh10.25BA.5B.7 C.8 D.1411.1101190079111141201222A.007110119114911120122 B.007110119114911122120C.007110911114119120122 D.11012091112211400711912.1.2 GHz4CPICPIA50%2B20%3C10%4D20%5MIPSA.100 B.200 C.400 D.60013.IEEE 754C640 0000H2 4 c=92 4 2 4 g=6a=3d=4e=6b=8h=9f=10a b c d e f g h A.-1.5213B.-1.5212C.-0.5x213D.-0.521214.8xyx=1 1110100y=10110000z=2*x+y/2zA.1 1000000 B.0 0100100 C.1 0101010 D.15.8/A.2B.3C.4D.516.256 MB4 GB4 KBTLB40FF180H0002H13FFF1H0035H002FF3H0351H103FFFH0153H03FF F180HA.015 3180HB.003 5180H C.TLBD.17.R1000H2000 H1000H2000H2000H3000H3000 H4000HA.1000H B.2000H C.3000H D.4000 H18.CPU1.03 GHz41CPU100A.0.25109/B.0.97109/C.1.0109/D.1.03 109/19.I/OA.PCI B.USB C.AGP D.PCI-Express20.RAIDI.II.III.IV.CacheA.III B.IIII C.IIIIIV D.IIIIIIV21.10 000/6 ms20 MB/s0.2 ms4 KBA.9 msB.9.4 ms C.12 ms D.12.4 ms22.I/ODMAA.I/OCPUDMAB.DMAC.I/ODMAD.I/ODMA23.A.B.C.D.24.CD-ROMA.B.C.D.25.I/OA.B.C.D.26.inodeA.B.C.D.27.1100151902A.200 B.295 C.300D.39028.I.II.sinIII.readA.III B.IIII C.IIIII D.IIIIII29.A.BIOSB.ROM C.EPROM D.RAM30.I.II.III.A.III B.IIIII C.IIII D.IIIIII31.P1P2P3CPUI/O 905100I/OP190%10%P250%50%P315%85%A.P1P2P3 B.P3P2P1C.P2P1=P3 D.P1P2=P332.A.B.C.D.33.OSIA.B.C.D.34.10 BaseTA.0011 0110B.1010 1101 C.0101 0010D.1100 010135.110 Mbps10 kb18 Mb1M=106.A.800 ms1 600 ms B.801 ms1 600 msC.1 600 ms800 ms D.1 600 ms801 ms36.A.CDMA B.CSMA C.TDMA D.FDMA37.HDLC01111100 01111110A.01111100 00111110 10 B.01111100 01111101 01111110C.01111100 01111101 0 D.01111100 01111110 0111110138.100Mbpscut-through switchingA.0 s B.0.48 s C.5.12 s D.121.44s39.TCP1TCP19132046100TCPA.20462012B.20462013 C.20472012 D.2047201340.SMTPI.7ASC IIII.III.IV.A.IIIIII B.IIIIVC.IIIIIV D.IIIIIIV41477041.13011(,)nAa aa0(0)ianin12pppmaaax/2(0,1)kmnpnkmxAA=(05535755)5A=(05535157)AAnA-112CC+Java342.104S=dofor repeat whilep1=0.35p2=0.15p3=0.15p4=0.35S42.21S2S43.932CPU800MHzCacheCPI4Cache3283240 ns32200 MHz32321CPU2Cache34BP1001.2Cache5%BPCPU44.1416CPU/CFZFNF00000OPCZNCFZFNF11C=1Z=0N=1CFNFCF=1NF=1OFFSETPC+2+2OFFSETPC+212200CHCF=0ZF=0NF=1PCCF=1ZF=0NF=0PC3CZN445.7500cobegini15 11 10 9 8 7 00 0 0 0 00111 1 1 0 0 0 1 115 11 10 9 8 7 00 0 0 0 0CZNOFFSET OP C Z N OFFSET PC2coendPVwaitsignal46.8324120122101012LA310000 8000H8 KB0090 0000H0020 0000H247.9Internet47ASIR1AS2R2R33 21 2 1 30090 0000H210020 0000HR2R1R3IP47471R2472R2IP194.17.20.200IPR2IP3R1R2153.14.5.0/25153.14.5.128/25AS1R1153.14.3.2194.17.20.0/25194.17.20.128/25AS2R2153.14.3.2194.17.21.0/24R3SOEOS1194.17.24.2 2013 1.Dmnmnmn2.C31n3.D7T034.B6(23)3(45)2(67)1465.AXX6.C7.C8.DD9.C5 2 1 4 6 7 3 AOE10.A25B52B11.C12C12.CCPI2 0.5 3 0.24 0.1 5 0.231.2GHa1200MHzMIPS1 200/3=40013.AIEEE 754C640 0000H1100 0110 0100 00000000 0000 0000 0000S11000 1100100 0000 0000 0000 0000 0000131.5 214.Axy1 1000000A15.Ckn21knk8n4k42(16)84 1(13)416.A03FF F180H03FFFH180H03FFFH0153H015 3180 H17.D4000H18.C 1000 H 2000 H 1000 H2000 H2000 H3000 H3000 H4000 H 41004(100 1)103CPU1.03 GHz1.03 G91.03G 100/1031.0 10/19.BUSBB20.BRAID21.B10 000/6 ms3 ms6 ms4 KB0.2 ms0.2 ms3+6+0.2+0.2=9.4 ms22.DI/OCPUCPUI/OCPUCPUDMADMACPUI/ODMACPUD23.A24.A25.CC26.AA27.C1105212105290300C28.BreadB29.DRAMD30.B31.BCPUI/OB32.BB33.BOSIOSIB34.A0011 0110A35.D8 Mb/10 Mb/s=800 ms800 ms1 600 ms10kb/10Mb/s=1 ms1 ms800801 ms36.BCSMAB37.AHDLCHDLC0111 1110510A38.B146 B0.48s39.B1TCPseq=1913ack=2046100TCPseq1=ack=2046ack1=seq+100=2013B40.ASMTPSMTP7ASC II 41.14NumNum NumcNum1Num110c1 ccn/227int Majority(int A,int n)int i,c,count=1;/ccountc=A0;/A0for(i=1;i 0)/count-;else /c=Ai;SMTPSMTPSMTPSMTPTCPPOP3POP3SMTPPOP3TCPTCPSMTPcount=1;if(count0)for(i=count=0;i n/2)return c;/else return-1;/121212O(n)O(1)47O(n)O(n)46Majority1O(nlog2n)36O(n2)35int Majority1(int A,int n)/O(n),O(n)int k,*p,max;p=(int*)malloc(sizeof(int)*n);/for(k=0;k n;k+)p k=0;/0max=0;for(k=0;kp max )max=Ak;/if(p max n/2)return max;else return-1;CC+JavaC32OnO1142.121=0.351+0.352+0.153+0.154=2.12221=0.351+0.352+0.153+0.154=2.1221=0.151+0.352+0.352+0.153=2.02122143.1CPU1/800 MHz=1.25 ns11/200 MHz=5 ns14 B200 MHz=800 MB/s4 B/5 ns=800 MB/s12Cache32 BCache1332 B1fordowhilerepeat1repeatdowhilefor240 ns/8=5 ns140 ns85 ns+40 ns+85 ns=85 ns24BPCPUCacheCache10041.25 ns=500 ns1Cache1.21005%85 ns=510 nsBPCPU500 ns+510 ns=1 010 ns2=CPI+44.116PC+21OFFSET8-1281271272OFFSET2C=0Z=1N=1ZFNFCF=0ZF=0NF=111110 0011B=E3HFFE3 H2FFC6 HPC200CH+2+FFC6H=1FD4H2CF=1ZF=0NF=01PC200CH+2=200EH13CZNC=Z=1N=034345.Semaphore empty=500;/2Semaphore mutex=1;/2i;P(empty);P(mutex);V(mutex);P(mutex);V(mutex);V(empty);coend3114 mutexPV2 emptyPV1 46.1124 KB1232/4K=2202204 B=4 MB22(unsigned int)(LA)22)&0 x3FF1(unsigned int)(LA)12)&0 x3FF1(unsigned int)(LA)22LA C310000 8000H888=+8=00200000H+84=0020 0020H351211213147.00901H00900H 2 1 0090 1000H0090 0000H0020 0024H0020 0020H16AS1153.14.5.0/25153.14.5.128/25153.14.5.0/24AS2194.17.20.0/25194.17.21.0/24194.17.20.0/23194.17.20.128/25194.17.20.128/25R2E0R2153.14.5.0/24153.14.3.2S0194.17.20.0/23194.17.24.2S1194.17.20.128/25E0 126IP0.50.5IP0.50.5 32IPIP194.17.20.200194.17.20.0/23194.17.20.128/25R2E01P13R1R2BGP4BGP1BGP4TCP1EGPEGPIP

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