膀胱肿瘤在病理上最重要的是（） A.组织类型 B.分化程度 C.病变部位 D.浸润深

After its (1) to the (2) in 2001, China has taken steps towards opening up its (3) . As a condition for joining the WTO, China (4) itself to a (5) of trade liberalization, including the reduction of tariffs on cars and agricultural products. These tariffs reductions are both (6) and there is (7) for China to (8) at a faster pace.
A (9) of (10) and a more rapid and profound reduction of import tariffs would (11) of those Chinese consumers who would (12) imported products at lower prices. Second, a higher (13) into China would take away much upward pressure on the Renminbi and provide a true and lasting (14) to solving current global (15) . Indeed, even a modest (16) in the opening-up process would go a long way towards solving the problem.
The opening up of China’s markets to international competition should (17) deep reforms. Banks should be recapitalized, bad loans provisioned fully and state-owned companies’ runaway indebtedness (18) . Only when real progress in these areas is made should China begin to consider a reform to its (19) . Doing it any earlier would be highly (20) .

患者，男，40岁，素有咳喘之疾，此次因起居不慎，感冒风寒，致头痛，发热，汗出，微恶风寒，同时伴有喘息发作，胸闷憋气，有少量白痰，舌淡红，苔薄白，脉浮。

其治法是（）。

A.发汗解表，宣肺平喘

B.辛温解表，温化水饮

C.清解肠热

D.清宣肺热

E.解肌祛风，降气平喘

下列属初步核实情况报告内容的有（）。

A、初核依据和初核工作概况

B、初核对象的简要情况

C、对所反映问题的初核情况

D、处理建议

下列市场调查的内容中，属于销售渠道调查的有( )。

A．现有渠道调查

B．中间商的销售情况分析

C．本企业销售策略变化后可能造成的销售变化

D．竞争产品的特性分析

E．消费者购买动机

下列给定程序中，函数fun的功能是：计算N×N矩阵的主对角线元素和反向对角线元素之和，并作为函数值返回。要求先累加主对角线元素中的值，再累加反向对角线元素中的值。
例如，若N=3，有下列矩阵：
1 2 3
4 5 6
7 8 9
首先累加1、5、9，然后累加3、5、7，函数返回值为30。
请在下划线处填入正确的内容并将下划线删除，使程序得出正确的结果。
注意：部分源程序给出如下。
不得增行或删行，也不得更改程序的结构!
试题程序：
#include＜stdio.h＞
#defineN4
fun(int t[][N],int n)

int i,sum;
/***********found***********/
______;
for(i=0;i＜n;i++)
/************found**********/
sum+=______;
for(i=0;i＜n;i++)
/**********found***********/
sum+=t[i][n-i-______];
return sum;

main()

int t[][N]=21,2,13,24,25,16,47,38,29,11,32,54,42,21,3,10,i,j;
printf("\nThe original data:\n");
for(i=0;i＜N;i++)

for(j=0;j＜N;j++)
printf("%4d",t[i][j]);
printf("\n");
printf("The result is:%d",fun(t,N));