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A【S1】svstem can be made up of many different functional activities，some of which are descr

ibed briefly below．Information links all areas of the logistics system together．Information processing is becoming increasdngly automated，complex，and rapid．It is key to the efficient functioning of system．【S2】is a broad area concerning all movements of raw materials，work in process，or finished goods within a factory or warehouse．【S3】processing is the system a firm has for getting orders from cuslomers，checking on the status cf orders and communicating to customers about them，and actually filling the order and making it available to the customer．【S4】involves selection of the mode,the routing of the shipment．compliance with regulations in the region of the country． and selection of the carriers．【S5】and storage activities relate lo warehouse layout，design，ownership，automation，training of employees，and relatted issues．

【S1】

A．management

B．logistics

C．information

D．shipping

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更多“A【S1】svstem can be made up of …”相关的问题

第1题

以下程序的输出结果是______ #includemain（）{char*s1,*s2,m;s1=s2=（char*）malloc（sizeof

以下程序的输出结果是______

# include

main（）

{char *s1,*s2,m;

s1=s2=（char*）malloc（sizeof（char））;

*s1=15;

*s2=20;

m=*s1+*s2;

printf（“%d\n”,m）;

}

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第2题

以下程序的输出结果是（)。 include＜stdlib.h＞ main（) {char*s1,*s2,m； s1=s2=（char*)malloc（size

以下程序的输出结果是()。

include＜stdlib.h＞

main()

{char*s1,*s2,m；

s1=s2=(char*)malloc(sizeof(char))；

*s1=15;

*s2=20;

m=*s1+*s2：

printf("%d\n",m)；

}

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第3题

Routing．protocols use different techniques for assigning【S1】to individual networks．Further

，each routing protocol forms a metric aggregation in a different．way．Most routing protocolscan use muh．iple paths if the paths have an equal【S2】．Some routing protocols can evenuse multiple paths when paths have an unequal cost．In either ease，load【S3】can improveoverall allocation of network bandwidth．When mulliple paths are used.there are several ways to distribute the packets．The two most common mechanisms are per—packet load balancing and per—destination load balancing．Per—packet load balancing distributes the【S4】across the possi-ble routes in a manner proportional to the route metrics．Per-destination load balancing distributes packets across the possible routes based on【S5】．

【S1】

A．calls

B．metrics

C．1inks

D．destinations

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第4题

Documentary credits are separate transactions from the sales contract with which they are

related and Letter of Credit can be divided into many kinds depending on the circumstance．A documentary credit may be available by payment，by negotiation or by acceptance．【S1】means that the nominated bank will pay the beneficiary the full amount due once he submits the contract documents required under the credit．Under a【S2】，the beneficiary is given double assurance of payment since the confirming bank has added its own undertaking to that of the opening bank．If a credit can be transferred by the original beneficiary to one or more parties，it is a【S3】．If a credit stipulated that its amount can be renewed without specific amendment to the credit being made，it is then a 【S4】．【S5】are those that cannot be amended or revoked without the consent of all the parties concerned．

【S1】

A．Confirmed credit

B．Payment credit

C．Negotiation credit

D．Acceptance credit

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第5题

执行以下程序后，test.txt文件的内容是(若文件能正常打开)______。 #include ＜stdio.h＞ main() { FILE *fp; char *s1="Fortran",*s2="Basic"; if((fp=fopen("test.txt","wb"))=NULL) { printf("Can't open test.txt file\n"); exit(1);} fwrite(s1,7,1,fp); /* 把从地址s1开始的7个字符写到fp所指文件中*/ f seek(fp, 0L,SEEK_SET);/*文件位置指针移到文件开头*/ fwrite(s2,5,1,fp); fclose(fp); }

A.Basican

B.BasicFortran

C.Basic

D.FortranBasic

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第6题

有以下程序：#include ＜iostream＞using namespace std;class sample{private: int n;public: samp

有以下程序： #include ＜iostream＞ using namespace std; class sample { private: int n; public: sample(){} sample (int m) { n=m; } sample add(sample s1,sample s2) { this-＞n=s1.n+s2.n; return (*this); } void disp() { cout＜＜"n="＜＜n＜＜end1; } }; int main () { sample s1(10) ,s2(5),s3; s3.add(s1,s2); s3.disp(); return 0; } 程序运行后，输出的结果是()。

A．n=10

B．n=5

C．n=20

D．n=15

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第7题

设25-75℃水的摩尔定压热容C_p,m为定值.恒定压力下将水由25℃加热到50℃的墒变为ΔS₁,由50℃加热到75℃的熵变为ΔS₂,则ΔS₁()ΔS₂.

设25-75℃水的摩尔定压热容C_p,m为定值.恒定压力下将水由25℃加热到50℃的墒变为ΔS₁,由50℃加热到75℃的熵变为ΔS₂,则ΔS₁（)ΔS₂.

A.＞

B.＜

C.=

D.≤

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第8题

执行以下程序后，test.txt文件的内容是（若文件能正常打开) （)。#include ＜stdio.h＞#include ＜stdli

执行以下程序后，test.txt文件的内容是(若文件能正常打开) ()。#include ＜stdio.h＞#include ＜stdlib.h＞main(){ FILE * fp; char * s1 = "Fortran" , * s2 = "Basic"; if((fp = fopen("test. txt" ," wb" )) = = NULL) { prinff("Can't open test. txt file \n"); exit(1); } fwrite(s1 ,7,1 ,fp); /* 把从地址s1开始到7个字符写到fp所指文件中*/ fseek(fp,OL,SEEK_SET); /*文件位置指针移到文件开头*/ fwrite (s2,5,1,fp); felose (fp);}

A．Basiean

B．BasieFortran

C．Basic

D．FortranBasie

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第9题

执行以下程序后， test.txt 文件的内容是（若文件能正常打开 )#include ＜stdio.h＞mai

执行以下程序后， test.txt 文件的内容是 (若文件能正常打开 )

#include ＜stdio.h＞

main()

{ FILE *fp;

char *s1="Fortran",*s2="Basic";

if((fp=fopen("test.txt","wb ” ))==NULL)

{ printf("Can't open test.txt file\n");exit(1);}

fwrite(s1,7,1,fp); /* 把从地址 s1 开始的 7 个字符写到 fp 所指文件中 */

fseek(fp,0L,SEEK_SET); /* 文件位置指针移到文件开头 */

fwrite(s2,5,1,fp);

fclose(fp);

}

A)Basican

B)BasicFortran

C)Basic

D)FortranBasic

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第10题

木材纤维细胞由初生壁P层和次生壁S层等构成。S层又可分为S1层、S2层和S3层，其中以S3层最厚，约为3～10μm。（）

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