Soluciones y Explicación de los problemas ACM ... - ICPC Bolivia

Explicaciones a los problemas South American Regionals 2008 19
int check(int x) {
int acc, i;
acc = 0;
for (i = 0; i < MAX; i++) {
acc /= x;
acc += lhs.c[i];
if (acc % x != 0) {
return 0;
}
}
return (acc == 0);
}
int main() {
int TC;
int h, i, j, max, fail;
char *p;
char buf[2*MAX+2];
for (;;) {
gets(buf);
if (buf[0] == ’=’) {
break;
}
max = 1;
for (p = buf; *p; p++) {
if (*p == ’=’) {
continue;
}
if (*p - ’0’ > max) {
max = *p - ’0’;
}
}
lhs = poly_read(buf);
for (p = buf; *p != ’=’; p++)
;
rhs = poly_read(++p);
for (i = 0; i < MAX; i++) {
lhs.c[i] -= rhs.c[i];
}
for (j = 0; j < MAX; j++) {
if (lhs.c[j]) break;
}
if (j == MAX) {
printf("%d+\n", max+1);
continue;
}