再藉著上界和下界把質數都輸出來。
P.S. 這題1也算是要輸出的東西之一。
[C](0.064)
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#include<stdio.h> | |
int main() | |
{ | |
int prime[1005] = { 1, 0}; | |
int prime_count[1005] = {0, 1}; | |
int count = 1; | |
int i, j; | |
for( i = 2 ; i <= 1000 ; i++ ) | |
{ | |
if( !prime[i] ) | |
{ | |
count++; | |
for( j = i+i ; j <= 1000 ; j += i ) | |
prime[j] = 1; | |
} | |
prime_count[i] = count; | |
} | |
int N, C; | |
while( scanf( "%d%d", &N, &C ) != EOF ) | |
{ | |
int start, end; | |
if( !( prime_count[N] % 2 ) ) | |
{ | |
start = prime_count[N]/2 - C + 1; | |
if( start < 1 ) | |
start = 1; | |
end = prime_count[N]/2 + C; | |
if( end > prime_count[N] ) | |
end = prime_count[N]; | |
} | |
else | |
{ | |
start = (prime_count[N]/2 + 1) - C + 1; | |
if( start < 1 ) | |
start = 1; | |
end = (prime_count[N]/2 + 1) + C - 1; | |
if( end > prime_count[N] ) | |
end = prime_count[N]; | |
} | |
printf( "%d %d:", N, C ); | |
count = 0; | |
for( i = 1 ; i <= N ; i++ ) | |
{ | |
if( !prime[i] ) | |
{ | |
count++; | |
if( count <= end && count >= start ) | |
printf( " %d", i ); | |
} | |
} | |
printf( "\n\n" ); | |
} | |
return 0; | |
} |
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