求N的所有因子(约数)

#include<stdio.h>
#include
<math.h>
#include
<string.h>
#include
<stdlib.h>
#define nmax 100000
int prime[nmax], flag[nmax], factor[nmax], cfactor[nmax], divisor[nmax];
int plen, flen, dlen;
void init() {
memset(flag,
-1, sizeof(flag));
int i, j;
for (i = 2, plen = 0; i < nmax; i++) {
if (flag[i]) {
prime[plen
++] = i;
}
for (j = 0; (j < plen) && (i * prime[j] < nmax); j++) {
flag[i
* prime[j]] = 0;
if (i % prime[j] == 0) {
break;
}
}
}
}
void findFactor(int n) {
int i, te, cnt;
te
= (int) (sqrt(n * 1.0));
for (i = 0, flen = 0; (i < plen) && (prime[i] <= te); i++) {
if (n % prime[i] == 0) {
cnt
= 0;
while (n % prime[i] == 0) {
cnt
++;
n
/= prime[i];
}
factor[flen]
= prime[i];
cfactor[flen]
= cnt;
flen
++;
}
}
if (n > 1) {
factor[flen]
= n;
cfactor[flen]
= 1;
flen
++;
}
}
void dfs(int k, int now) {
if (k == flen) {
divisor[dlen
++] = now;
return;
}
int i;
for (i = 0; i < cfactor[k]; i++) {
now
= now * factor[k];
dfs(k
+ 1, now);
}
for (i = 0; i < cfactor[k]; i++) {
now
= now / factor[k];
}
dfs(k
+ 1, now);
}
int cmp(const void *a, const void *b) {
return *(int *) a - *(int *) b;
}
int main() {
#ifndef ONLINE_JUDGE
freopen(
"data.in", "r", stdin);
#endif
init();
int n, i;
while (~scanf("%d", &n)) {
findFactor(n);
dlen
= 0;
dfs(
0, 1);
qsort(divisor, dlen,
sizeof(divisor[0]), cmp);
for (i = 0; i < dlen; i++) {
printf(
"%d ", divisor[i]);
}
printf(
"\n");
}
return 0;
}

另一种方法为:

#define nmax 31625
#define fac 50005
map<int, int> mymap;
int prime[nmax], plen, factor[fac], pfactor[fac], cpfactor[fac], len_pfactor,
len_factor;
void dfs(int p, int c, long long nn, int n) {
int i, r;
long long te;
for (i = p; i < len_pfactor; i++) {
if (c > cpfactor[i]) {
dfs(i + 1, 0, nn, n);
}
te = nn * pfactor[i];
if (te >= n) {
return;
}
r = (int) te;
if (n % r == 0) {
if (!mymap[r]) {
factor[len_factor++] = r;
mymap[r] = 1;
}
}
dfs(i, c + 1, te, n);
}
}
solve(n);
factor[0] = 1;
len_factor = 1;
mymap.clear();
dfs(0, 0, 1, n);
if (n != 1) {
factor[len_factor++] = n;
}



posted @ 2011-09-20 18:07  qingyezhu  阅读(294)  评论(0编辑  收藏  举报