华东师范大学计算机机试真题.docx

1、华东师范大学计算机机试真题华东师范大学计算机机试真题2009机试计算和的数位Sum of digit Description Write a program which computes the digit number of sum of two integers a and b. Input The first line of input gives the number of cases, N(1 N 100). N test cases follow.Each test case consists of two integers a and b which are separeted

2、by a space in a line. (0=a,b=100000000). Output For each test case, print the number of digits of a + b. Sample Input 35 71 991000 999 Sample Output 234 #includeint main()int n;int a,b;int sum;while(scanf(%d,&n)!=EOF)while(n-) int an=0; scanf(%d%d,&a,&b); sum=a+b; while(sum) an+; sum/=10; printf(%dn

3、,an+);return 0;大写改小写Capitalize Description Write a program which replace all the lower-case letters of a given text with the corresponding captital letters. Input A text including lower-case letters, periods, and space.Output Output The converted text. Sample Input welcome to east china normal unive

4、rsity. Sample Output WELCOME TO EAST CHINA NORMAL UNIVERSITY. #include#includechar str1000;int main()int l;while(gets(str)l=strlen(str);int i;for(i=0;i=a&stri=z)printf(%c,stri-32);elseprintf(%c,stri);printf(n);return 0;素数对Primes Pair Description We arrange the numbers between 1 and N (1 = N = 10000)

5、 in increasing order and decreasing order like this: 1 2 3 4 5 6 7 8 9 . . . NN . . . 9 8 7 6 5 4 3 2 1Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both numbers in the pairs are prime. Input The first line of input gives the number of cases, C (1

6、C 100). C test cases follow.Each test case consists of an integer N in one line. Output For each test case, output P . Sample Input 414751 Sample Output 0226 #include#includebool prime10005;void init()int i;int j;prime0=prime1=false;/不是素数prime2=true;/是素数for(i=3;i=10005;i+=2)primei=true;/是素数primei+1=

7、false;/不是素数 除0和2之外的偶数都不是素数for(i=3;i=10005;i+=2)if(primei=true)/是素数j=i+i;while(j=10005)primej=false;/不是素数j+=i;int main()int c;int n;init();/初始化while(scanf(%d,&c)!=EOF)while(c-)scanf(%d,&n);int sum=0;int i;for(i=2;i=n/2;i+)if(primei=true&primen+1-i=true)sum+;sum*=2;if(n%2=1)/n为奇数if(primen/2+1=true)sum

8、+=1;printf(%dn,sum);return 0;求最大公约数和最小公倍数GCD and LCM Description Write a program which computes the greatest common divisor (GCD) and the least common multiple (LCM) of given a and b (0 a, b 44000). Input The first line of input gives the number of cases, N(1 N 100). N test cases follow.Each test ca

9、se contains two interger a and b separated by a single space in a line. Output For each test case, print GCD and LCM separated by a single space in a line. Sample Input 28 65000 3000 Sample Output 2 241000 15000 #includeint getgcd(int a,int b)int gcd;int t1,t2;t1=a;t2=b;gcd=t1%t2;while(gcd!=0)t1=t2;

10、t2=gcd;gcd=t1%t2;return t2;int main()int n;int a,b;while(scanf(%d,&n)!=EOF)while(n-)scanf(%d%d,&a,&b);printf(%d %dn,getgcd(a,b),a*b/(getgcd(a,b);return 0;排序后求位置处的数Sort it Description There is a database,partychen want you to sort the databases data in the order from the least up to the greatest elem

11、ent,then do the query: Which element is i-th by its value?- with i being a natural number in a range from 1 to N.It should be able to process quickly queries like this. Input The standard input of the problem consists of two parts. At first, a database is written, and then theres a sequence of queri

12、es. The format of database is very simple: in the first line theres a number N (1=N=100000), in the next N lines there are numbers of the database one in each line in an arbitrary order. A sequence of queries is written simply as well: in the first line of the sequence a number of queries K (1 = K =

13、 100) is written, and in the next K lines there are queries one in each line. The query Which element is i-th by its value? is coded by the number i. Output The output should consist of K lines. In each line there should be an answer to the corresponding query. The answer to the query i is an elemen

14、t from the database, which is i-th by its value (in the order from the least up to the greatest element). Sample Input 5712112371213325 Sample Output 1217123#include#includeusing namespace std;int num100010;int pos105;int main()int n;int i;int k;while(scanf(%d,&n)!=EOF)for(i=1;i=n;i+)scanf(%d,&numi);scanf(%d,&k);for(i=1;i=k;i+