uva 254 - Towers of Hanoi(递归）

题目链接：uva 254 - Towers of Hanoi

题目大意：给出n和k，n的范围为0~100， k的范围为0~2^n - 1， 然后模拟汉诺塔游戏移动n个碟子，按照最优的游戏策略，移动了k次后，三个柱子上各有多少个碟子。

注意：如果盘子的总数为奇数，那么移动的目标最终目标位B号柱，否则为C号柱（题目要按照某种序列移动碟子）。

解题思路：递归的思想，每次处理好说当前移动碟子所在的柱子，目标柱子以及中间辅助的柱子，然后枚举当前的步数够移动几个碟子。k很大，用double和long long处理不了，就写了大数。

#include <stdio.h>
#include <string.h>
#include <math.h>
#define max(a,b) (a)>(b)?(a):(b)
const int N = 1005;

struct bign {
int s
;
bign () {
memset(s, 0, sizeof(s));
}
bign (int number) {*this = number;}
bign (const char* number) {*this = number;}

void put();
void del();

bign operator = (char *num) {
s[0] = strlen(num);
for (int i = 1; i <= s[0]; i++)
s[i] = num[s[0] - i] - '0';
return *this;
}

bign operator = (int num) {
char str
;
sprintf(str, "%d", num);
return *this = str;
}

bool operator < (const bign& b) const {
if (s[0] != b.s[0])
return s[0] < b.s[0];
for (int i = s[0]; i; i--)
if (s[i] != b.s[i])
return s[i] < b.s[i];
return false;
}
bool operator > (const bign& b) const { return b < *this; }
bool operator <= (const bign& b) const { return !(b < *this); }
bool operator >= (const bign& b) const { return !(*this < b); }
bool operator != (const bign& b) const { return b < *this || *this < b;}
bool operator == (const bign& b) const { return !(b != *this); }

bign operator + (const bign& c) {
int sum = 0;
bign ans;
ans.s[0] = max(s[0], c.s[0]);

for (int i = 1; i <= ans.s[0]; i++) {
if (i <= s[0]) sum += s[i];
if (i <= c.s[0]) sum += c.s[i];
ans.s[i] = sum % 10;
sum /= 10;
}

while (sum) {
ans.s[++ans.s[0]] = sum % 10;
sum /= 10;
}
return ans;
}

bign operator - (bign& c) {
bign ans = *this;
for (int i = 1; i <= c.s[0]; i++) {
if (ans.s[i] < c.s[i]) {
ans.s[i] += 10;
ans.s[i + 1] -= 1;
}
ans.s[i] -= c.s[i];
}

ans.del();
return ans;
}
};

bign cnt[105], tmp = 1, over = 0;

void init() {
cnt[0] = 0;
for (int i = 1; i <= 100; i++)
cnt[i] = cnt[i - 1] + cnt[i - 1] + tmp;
}

void hanoi(int n, bign k, int& a, int& b, int& c) {
if (k == over) {
a += n;
return;
}

for (int i = 1; i <= n; i++) {
if (cnt[i] == k) {
a += n - i;
if (n % 2 == i % 2)
c += i;
else
b += i;
return;
} else if (cnt[i] > k) {
a += n - i;
if (n % 2 == i % 2) {
c += 1;
hanoi(i - 1, k - cnt[i - 1] - tmp, b, a, c);
} else {
b += 1;
hanoi(i - 1, k - cnt[i - 1] - tmp, c, a, b);
}
return;
}
}
}

int main () {
init();
int n, a, b, c;
bign k;
char str
;
while (scanf("%d%s", &n, str), n || strcmp(str, "0")) {
a = b = c = 0;
k = str;

if (n % 2)
hanoi(n, k, a, c, b);
else
hanoi(n, k, a, b, c);
printf("%d %d %d\n", a, b, c);
}
return 0;
}

void bign::put() {
if (s[0] == 0)
printf("0");
else
for (int i = s[0]; i; i--)
printf("%d", s[i]);
}

void bign::del() {
while (s[s[0]] == 0) {
s[0]--;
if (s[0] == 0) break;
}
if (s[0] == 0) {
s[0] = 1;
s[1] = 0;
}
}