hdu 5429 Geometric Progression(高精度)

解题思路

代码

#include <cstdio>
#include <cstring>
#include <iostream>

using namespace std;
const int maxn = 10005;

struct bign {
int len, num[maxn];

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

void delzero ();
void Put ();

void operator = (int number);
void operator = (char* number);

bool operator <  (const bign& b) const;
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); }

void operator ++ ();
void operator -- ();
bign operator + (const int& b);
bign operator + (const bign& b);
bign operator - (const int& b);
bign operator - (const bign& b);
bign operator * (const int& b);
bign operator * (const bign& b);
bign operator / (const int& b);
//bign operator / (const bign& b);
int operator % (const int& b);
};

const int maxm = 105;

int N;
char s[maxm];
bign A[maxm], zero = 0;

bool judge () {
int c = 0;
for (int i = 0; i < N; i++) if (A[i] == zero) c++;
if (c) return c == N;
for (int i = 1; i < N-1; i++)
if (A[i-1] * A[i+1] != A[i] * A[i]) return false;
return true;
}

/*Code*/
int main () {
int cas;
scanf("%d", &cas);
while (cas--) {
scanf("%d", &N);
for (int i = 0; i < N; i++) {
scanf("%s", s);
A[i] = s;
}
printf("%s\n", judge() ? "Yes" : "No");
}
return 0;
}

void bign::delzero () {
while (len && num[len-1] == 0)
len--;

if (len == 0) {
num[len++] = 0;
}
}

void bign::Put () {
for (int i = len-1; i >= 0; i--)
printf("%d", num[i]);
}

void bign::operator = (char* number) {
len = strlen (number);
for (int i = 0; i < len; i++)
num[i] = number[len-i-1] - '0';

delzero ();
}

void bign::operator = (int number) {

len = 0;
while (number) {
num[len++] = number%10;
number /= 10;
}

delzero ();
}

bool bign::operator < (const bign& b) const {
if (len != b.len)
return len < b.len;
for (int i = len-1; i >= 0; i--)
if (num[i] != b.num[i])
return num[i] < b.num[i];
return false;
}

void bign::operator ++ () {
int s = 1;

for (int i = 0; i < len; i++) {
s = s + num[i];
num[i] = s % 10;
s /= 10;
if (!s) break;
}

while (s) {
num[len++] = s%10;
s /= 10;
}
}

void bign::operator -- () {
if (num[0] == 0 && len == 1) return;

int s = -1;
for (int i = 0; i < len; i++) {
s = s + num[i];
num[i] = (s + 10) % 10;
if (s >= 0) break;
}
delzero ();
}

bign bign::operator + (const int& b) {
bign a = b;
return *this + a;
}

bign bign::operator + (const bign& b) {
int bignSum = 0;
bign ans;

for (int i = 0; i < len || i < b.len; i++) {
if (i < len) bignSum += num[i];
if (i < b.len) bignSum += b.num[i];

ans.num[ans.len++] = bignSum % 10;
bignSum /= 10;
}

while (bignSum) {
ans.num[ans.len++] = bignSum % 10;
bignSum /= 10;
}

return ans;
}

bign bign::operator - (const int& b) {
bign a = b;
return *this - a;
}

bign bign::operator - (const bign& b) {
int bignSub = 0;
bign ans;
for (int i = 0; i < len || i < b.len; i++) {
bignSub += num[i];
bignSub -= b.num[i];
ans.num[ans.len++] = (bignSub + 10) % 10;
if (bignSub < 0) bignSub = -1;
else bignSub = 0;
}
ans.delzero ();
return ans;
}

bign bign::operator * (const int& b) {
int bignSum = 0;
bign ans;

ans.len = len;
for (int i = 0; i < len; i++) {
bignSum += num[i] * b;
ans.num[i] = bignSum % 10;
bignSum /= 10;
}

while (bignSum) {
ans.num[ans.len++] = bignSum % 10;
bignSum /= 10;
}

return ans;
}

bign bign::operator * (const bign& b) {
bign ans;
ans.len = 0;

for (int i = 0; i < len; i++){
int bignSum = 0;

for (int j = 0; j < b.len; j++){
bignSum += num[i] * b.num[j] + ans.num[i+j];
ans.num[i+j] = bignSum % 10;
bignSum /= 10;
}
ans.len = i + b.len;

while (bignSum){
ans.num[ans.len++] = bignSum % 10;
bignSum /= 10;
}
}
return ans;
}

bign bign::operator / (const int& b) {

bign ans;

int s = 0;
for (int i = len-1; i >= 0; i--) {
s = s * 10 + num[i];
ans.num[i] = s/b;
s %= b;
}

ans.len = len;
ans.delzero ();
return ans;
}

int bign::operator % (const int& b) {

bign ans;

int s = 0;
for (int i = len-1; i >= 0; i--) {
s = s * 10 + num[i];
ans.num[i] = s/b;
s %= b;
}

return s;
}