#include<stdio.h>
#include<math.h>
#include<algorithm>

using namespace std;

#define PI 3.1415926535897932384626433

int lat1, lat2, lat3, lon1, lon2, lon3, trash;
char latc, lonc;

long double latf, lats, lonf, lons;
long double sq(double x){
  return x*x;
}

int main(void){
  scanf("Message #%d.\nReceived at %d:%d:%d.\nCurrent ship’s coordinates are\n%d^%d'%d\" %cL\nand %d^%d'%d\" %cL.\n",
         &trash, &trash, &trash, &trash, &lat1, &lat2, &lat3, &latc, &lon1, &lon2, &lon3, &lonc);
  lats = (latc=='N'?1:-1)*(lat1 + (lat2 + lat3/60.0)/60.0)*PI/180.0;
  lons = (lonc=='W'?-1:1)*(lon1 + (lon2 + lon3/60.0)/60.0)*PI/180.0;


  scanf("An iceberg was noticed at\n%d^%d'%d\" %cL\nand %d^%d'%d\" %cL.",
         &lat1, &lat2, &lat3, &latc, &lon1, &lon2, &lon3, &lonc);
  latf = (latc=='N'?1:-1)*(lat1 + (lat2 + lat3/60.0)/60.0)*PI/180.0;
  lonf = (lonc=='W'?-1:1)*(lon1 + (lon2 + lon3/60.0)/60.0)*PI/180.0;

  double dlon = lonf - lons, dlat = latf - lats;

  double a = acos(sin(lats)*sin(latf) + cos(lats)*cos(latf)*cos(dlon));
  double len =  3437.5*a;
  printf("The distance to the iceberg: %.2lf miles.\n", len);
  if(100.00-len>0.005)
    printf("DANGER!\n");
  return 0;
}

Am I wright about input?

Have you debugged locally? Change the word "ship’s" to "ship's" and you'll get AC.
Anyway, your algorithm is amazing. For example, you even don't use epsilon in "if(100.00-len>0.005)"... Very impressive.