Pagini recente » Cod sursa (job #1545828) | Cod sursa (job #1773324) | Cod sursa (job #432786) | Cod sursa (job #898145) | Cod sursa (job #1818121)
#include <stdio.h>
#include <string.h>
using namespace std;
int f[2000001];
int n, m, p=2, k, p1, p2, nr;
int sol[1001];
char a[2000001], b[2000001];
int main()
{
freopen("strmatch.in","r",stdin);
freopen("strmatch.out","w",stdout);
gets(a), gets(b);
n = strlen(a), m = strlen(b);
f[0] = -1, f[1] = 0;
while (p < n)
if (a[p-1] == a[k]) f[p++] = ++k;
else
if (k>0) k = f[k];
else f[p++] = 0;
while (p1 + p2 < m)
if (a[p2] == b[p1 + p2])
if (p2 == n - 1) {
if (++nr <= 1000) sol[nr-1] = p1;
if (f[p2] > -1) p1 += p2 - f[p2], p2 = f[p2];
else p2 = 0, p1++;
}
else p2++;
else
if (f[p2] > -1) p1 += p2 - f[p2], p2 = f[p2];
else p2 = 0, p1++;
printf("%d\n",nr);
for (int i = 0; i < (nr<1000)?nr:1000; i++) printf("%d ",sol[i]);
}