The wholesaler buys goods in large quantities from the manufacturers and sells them in smaller parcels to retailers, and for this service his selling price to the retailer is raised several percent higher. But his job is made more difficult by retail demand not necessarily running level with manufacturers' production. Because he adjusts or regulates the flow of goods by holding stock until required, he frees the manufacturer, to some extent, from the effect on production of changing demand and having to bear the whole risk.
The manufacturer can then keep up a steady production flow, and the retailer has no need to hold heavy stocks, who can call on the wholesaler for supplies any time. This wholesale function is like that of a valve in a water pipe. The middleman also bears part of the risk that would otherwise fall on the manufacturer and also the retailer.
The wholesaler provides a purely commercial service, for which he is too well rewarded. But the point that is missed by many people is that the wholesaler is not just someone adding to the cost of goods. It is true one could eliminate the wholesaler but one would still be left with his function: that of making sure that goods find their way to the people who want them.
"Middleman" in the passage almost equals to all the following in meaning EXCEPT______.
A.go-between
B.intermediary
C.manufacturer
D.wholesaler
第1题
#include
#include
int fun(int *s, int t, int *k)
{ int i;
*k=0;
for(i=0;i
if(s[*k]
return s[*k]; }
main()
{ int a[10]={ 876,675,896,101,301,401,980,431,451,777},k;
fun(a, 10, &k);
printf("%d, %d\n",k,a[k]);}
如果输入如下整数:876 675 896 101 301 401 980 431 451 777
则输出结果为
A.7,431
B.6
C.980
D.6,980
第2题
A.7,431
B.6
C.980
D.6,980
第3题
A.7,431
B.6
C.980
D.6,980
第4题
请编写一个函数int fun(int *s,int t,int *k),用来求出数组的最大元素在数组中的下标并存放在k所指的存储单元中。
例如,输入如下整数:
876 675 896 101 301 401 980 431 451 777
则输出结果为6,980。
注意:部分源程序给出如下。
请勿改动主函数main和其他函数中的任何内容,仅在函数fun的花括号中填入所编写的若干语句。
试题程序:
include <conio.h>
include <stdio.h>
int fun(int *s,int t,int *k)
{
}
main()
{
int a[10]={ 876,675,896,101,301,401,
980,431,451,777},k;
clrscr();
fun(a, 10, &k);
printf("%d, %d\n ", k, a[k]);
}
第5题
下列程序是用来判断数组中特定元素的位置所在,则输出结果为 #include<conio.h> #include<iostream.h> int fun(int *s,int t,int*k) { int i; *k=0; for(i=0;i
A.7,431
B.6
C.980
D.6,980
第6题
请编写函数fun,其功能是求出数组的最大元素在数组中的下标并存放在k所指的存储单元中。
例如,输入如下整数:876 675 896 10l 30l 401 980 431451 777则输出结果为:6,980
注意:部分源程序在文件PROGl.C文件中。
请勿改动主函数main和其他函数中的任何内容,仅在函数fun的花括号中填入你编写的若干语句。
第7题
下列程序是用来判断数组中特定元素的位置所在,则输出结果为 #include<conio.h> #include<iostream.h> in[fun(int * s,int t,int * k) { int i; *k=0; for(i=0;<t;i++) if(s[*k]<s[i]) *k=i; return s[*k];} void main() { int a[10]={876,675,896,101,301,401,980,431,451,777},k; fun(a,10,&k); cout < < k < <‘,’ < < a[k];}
A.7,431
B.6
C.980
D.6,980
第8题
下列程序用来判断数组中特定元素的位置所在,则输出结果为()。 #include<conio.h> #include<iostream.h> int fun(int*s,int t,int*k) {int i; *k=0; for(i=0;i<t;i++) if (s[*k]<s[i])*k=i; return s[*k];} void main() {int a[10]=(876,675,896,101,301,401,980,431,451,777);int k; fun(a,10,&k); cout<<k<<','<<a[k];}
A.7,431
B.6
C.980
D.6,980
第9题
下列程序用来判断数组中特定元素的位置所在,则输出结果为()。 #include<conio.h> #include<iostream.h> int fun(int*p,int n,int*j) {int i; *j=0; for(i=0;i<n;i++) if(p[*j]<p[i])*j=i; return p[*j];} void main() {int a[10]={1,3,9,0,8,7,6,5,4,2)j;int j; fun(a,10,&j); cout<<j<<','<<a[j];}
A.2,9
B.3
C.9
D.3,9
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