阅读理解。

     Xi'an holds the International Horticultural Exposition, 2011. The International Horticultural Exposition(世园会)  has a long history and it'll show us the beauty of the nature. So the International Horticultural

Exposition is an honor for all of the Xi'an people. Our government has

promised that it will be the best one. And Xi'an, as a host city, had more chances to develop quickly. As

a student in Xi'an, I should learn English well so that I can be a volunteer in the exposition to help foreigners know more about Xi'an.

    The Expo emblem(徽章)  looks like the famous City Wall. The Expo is based on a general pattern of "

two circles,  two axes,  and five clusters. . . " Xi'an Horticultural Expo was held from April to October,

2011.

    Great changes have taken place in Xi'an and more and more people throughout the world are focusing

their attention on Xi'an now.  As a citizen of Xi'an, I must spare no efforts to do what I should do. First, I

decide to plant more trees to make our city more beautiful.  Second, I should obey seven nos and be good at learning from others.Third,I should study hard so that I can do something better for my city in the future.

1._______ is the host of the International Horticultural Exposition,2011.

2.  We students should learn English well in order that we can be a volunteer to help _______ know more about Xi'an.

3.  International Horticultural Exposition, 2011, Xi'an will last_______.

4. The Xi'an Horticultural Expo is based on a_______ of "two circles, two axes, and five clusters."

5. What can we do for the International Horticultural Expo?

    We should _______, obey seven nos, learn from others and work hard.

给定程序中，函数fun的功能是：将形参s所指字符串中的所有数字字符顺序前移，其他字符顺序后移，处理后新字符串的首地址作为函数值返回。
请在程序的下划线处填入正确的内容并把下划线删除，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＞
#include ＜string.h＞
#include ＜stdlib.h＜
#include ＜ctype.h＜
char *fun(char *s)
inti, j, k, n; char *p, *t;
n=strlen(s)+1;
t=(c har*)malloc (n*sizeof(char));
p=(char*)malloc(n*sizeof(char));
j=0; k=0;
for(i=0; i＜n; i++)
if(isdigit(s[i]))
/******************found*******************/
p[ (1) ]=s[i]; j++;
else
t[k]=s[i]; k++;

/******************found*******************/
for(i=0; i＜ (2) ; i++)p[j+i]=t[i];
p[j+k]=0;
/******************found*******************/
return (3) ;

main()

char s[80];
printf("Please input:");
scanf("%s", s);
printf("\nThe result is: %s\n", fun(s));