Description
A prime number is a counting number (1, 2, 3, ...)
that is evenly divisible only by 1 and itself. In this problem you are to write
a program that will cut some number of prime numbers from the list of prime
numbers between (and including) 1 and N. Your program will read in a number N;
determine the list of prime numbers between 1 and N; and print the C*2 prime
numbers from the center of the list if there are an even number of prime numbers
or (C*2)-1 prime numbers from the center of the list if there are an odd number
of prime numbers in the list.
Input
Each input set will be on a line by itself and will
consist of 2 numbers. The first number (1 <= N <= 1000) is the maximum
number in the complete list of prime numbers between 1 and N. The second number
(1 <= C <= N) defines the C*2 prime numbers to be printed from the center
of the list if the length of the list is even; or the (C*2)-1 numbers to be
printed from the center of the list if the length of the list is odd.
Output
For each input set, you should print the number N
beginning in column 1 followed by a space, then by the number C, then by a colon
(:), and then by the center numbers from the list of prime numbers as defined
above. If the size of the center list exceeds the limits of the list of prime
numbers between 1 and N, the list of prime numbers between 1 and N (inclusive)
should be printed. Each number from the center of the list should be preceded by
exactly one blank. Each line of output should be followed by a blank line.
Hence, your output should follow the exact format shown in the sample output.
Sample Input
21 2
18 2
18 18
100 7
Sample Output
21 2: 5 7 11
18 2: 3 5 7 11
18 18: 1 2 3 5 7 11 13 17
100 7: 13 17 19 23 29 31 37 41 43 47 53 59 61 67
我的程序，，自己测试时没问题的啊，怎么OJ一直报错
#include
#include
main()
{
int b,waitj,m=0,i;
int fuzhu=0;
int s;
int c;
int al;
int sushugeshu;
int w;
int maxxiabiao;
int maxnum,k,temp,pn[500];
system("cls");
scanf("%d%d",&maxnum,&c);
temp=maxnum;
for(k=0;k
temp=temp-1;
s=waitj-1;
for(i=1;i<=waitj-2;i++)
{
b=waitj%s;
s=s-1;
if(b==0)
goto ends;
}
pn[m]=waitj;
m=m+1;
ends:
printf("");
}
maxxiabiao=m-1;//m在pn[]赋值后又增大了1，故，pn最大下标应为m-1；现在令maxxiabiao为最大下标值
sushugeshu=m;//素数个数应当为下表+1，故sushugeshu=m
//printf("%d %d",sushugeshu,maxxiabiao);
if(m%2==0){
printf("%d %d:",maxnum,c);
if(c>(m/2))
c=(m/2);
for(w=m/2-c;w
fuzhu++;}}
if(m%2!=0){
printf("%d %d:",maxnum,c);
if(c>(m/2)+1)
c=(m/2)+1;
for(w=(m-1)/2-c+1;w<=(m-1)/2+c-1;w++){
printf(" %d",pn[(m-1)/2+c-1-fuzhu]);
fuzhu++;}}
error:
return 0;
}