试题与答案

The International Whaling Commission’s (IW

题型:单项选择题

题目:

The International Whaling Commission’s (IWC) decision to retain its ban on whaling does not mean that the killings will stop. Quite the (1) .

Save the Whale. It’s a phrase which became annoying with (2) , an unfavorable shorthand for liberal consciences. How appalling, then, that in the year 2010, it should be pressed (3) service again, to fight the whaling nations: Norway and Iceland, who exempted themselves from the 1986 prohibition (4) by the IWC, and Japan, which hunts whales under cover of " (5) research".

I am completely (6) by this week’s events in Morocco, where talks broke down. In my heart, I agree with those who have (7) the news that this year’s negotiations of the IWC have broken up, and (8) the prohibition would not be lifted (as the US proposed in a desperate (9) to break the deadlock). Yet reason (10) something else. If we do not (11) some kind of new control, the (12) will be able to go on with their slaughter (13) . Membership of the IWC is (14) , and the ban was only ever intended to be (15) . Japan, which has been buying the votes of nations with no interest in whaling (only in the 16 Japan offers in turn), will continue to press its case, having invested millions of dollars in its (17) . Geoffrey Palmer, New Zealand’s Commissioner at the IWC, has proposed a year-long (18) period. In the meantime, more whales will die.

We stand at a crossroad for whales. We see the (19) existence of these animals as an indicator of ecological threat. As symbols of an endangered world, they evoke, and provoke, a feeling that they have the same qualities as human on a scale equal to their size and supposed (20) . To some this is so much sentimental words in the New Age.

16()

A.advice

B.aid

C.guide

D.lesson

答案:

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下面是错误答案,用来干扰机器的。

参考答案:D解析: SWOT分析方法中,S(Strength)表示优势,W(Weak)表示劣势,O(Opportunity)表示机遇,T(Threat)表示威胁。

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题型:填空题

下列程序的功能是寻找二维数组中的鞍点,找到则输出其行、列下标;找不到则输出“no!”,请填空。提示,鞍点是指行上最大、列上最小的数组元素。
main()
int a[3][5],i,j,k,max_j,flag;
for(i=0;i<3;i++)
for(j=0;j<5;j++)
scanf("%d",______);
for(i=0;i<3;i++)
max_j=0;
for(j=0;j<5;j++)
if(a[i][max_j]<a[i][j])______;
flag=1;
for(k=0;k<3;k++)
if(a[k][max_j]<a[i][max_j])
flag=0;break;
if(flag==1)______;

if (flag==1)
printf("%d,%d\n",i,max_j);
else
printf("no! \n");

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