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for each given N find it's prime factors problem propsed in scorify challange .
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#include <stdio.h> | |
#include <string.h> | |
#include <math.h> | |
#include <stdlib.h> | |
int nextInt(){ | |
int a; scanf("%d", &a); return a; | |
} | |
long long nextLong(){ | |
long long a; scanf("%lld", &a); return a; | |
} | |
int m, n; | |
const int N = 1e5 + 10; | |
int main() { | |
int seive[N]; | |
int list[1000]; | |
/* the pre processing part */ | |
seive[1] = 1; | |
for (int i = 2; i < N; i++) | |
seive[i] = i; | |
for (int i = 4; i < N; i += 2) | |
seive[i] = 2; | |
for (int i = 3; i * i < N; i+=2) { | |
if (seive[i] == i) { | |
for (int j = i * i; j < N; j += i) | |
if (seive[j] == j) | |
seive[j] = i; | |
} | |
} | |
int t = nextInt(); | |
/* answer queries in O(log(N)) per query */ | |
for (int tc = 1; tc <= t; tc++){ | |
n = nextInt(); | |
int j = 0; | |
while (n != 1) { | |
list[j++] = (seive[n]); | |
n = n / seive[n]; | |
} | |
for(int i = 0; i < j; i++){ | |
if(i > 0) printf(" "); | |
printf("%d", list[i]); | |
} | |
printf("\n"); | |
} | |
return 0; | |
} |
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