UVa 10494 If We Were a Child Again (高精度)

10494 - If We Were a Child Again

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=97&page=show_problem&problem=1435

“Oooooooooooooooh!

If I could do the easy mathematics like my school days!!

I can guarantee, that I’d not make any mistake this time!!”

Says a smart university student!!

But his teacher even smarter – “Ok! I’d assign you such projects in your software lab. Don’t be so sad.”

“Really!!” - the students feels happy. And he feels so happy that he cannot see the smile in his teacher’s face.

 

 

The Problem

 

The first project for the poor student was to make a calculator that can just perform the basic arithmetic operations.

 

But like many other university students he doesn’t like to do any project by himself. He just wants to collect programs from here and there. As you are a friend of him, he asks you to write the program. But, you are also intelligent enough to tackle this kind of people. You agreed to write only the (integer) division and mod (% in C/C++) operations for him.

 

Input

Input is a sequence of lines. Each line will contain an input number. One or more spaces. A sign (division or mod). Again spaces. And another input number. Both the input numbers are non-negative integer. The first one may be arbitrarily long. The second number n will be in the range (0 < n < 2³¹).

 

 

Output

A line for each input, each containing an integer. See the sample input and output. Output should not contain any extra space.

 

 

 

Sample Input

110 / 100

99 % 10

2147483647 / 2147483647

2147483646 % 2147483647

 

 

 

 

 

 

Sample Output

1

9

1

2147483646

完整代码：

/*0.019s*/#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 1000;///不要开太小了！char numstr[maxn], numstr2[maxn];struct bign{int len, s[maxn];bign(){memset(s, 0, sizeof(s));len = 1;}bign(int num){*this = num;}bign(const char* num){*this = num;}bign operator = (const int num){char s[maxn];sprintf(s, "%d", num);*this = s;return *this;}bign operator = (const char* num){len = strlen(num);for (int i = 0; i < len; i++) s[i] = num[len - i - 1] & 15;return *this;}///输出const char* str() const{if (len){for (int i = 0; i < len; i++)numstr[i] = '0' + s[len - i - 1];numstr[len] = '\0';}else strcpy(numstr, "0");return numstr;}///去前导零void clean(){while (len > 1 && !s[len - 1]) len--;}///加bign operator + (const bign& b) const{bign c;c.len = 0;for (int i = 0, g = 0; g || i < max(len, b.len); i++){int x = g;if (i < len) x += s[i];if (i < b.len) x += b.s[i];c.s[c.len++] = x % 10;g = x / 10;}return c;}///减bign operator - (const bign& b) const{bign c;c.len = 0;for (int i = 0, g = 0; i < len; i++){int x = s[i] - g;if (i < b.len) x -= b.s[i];if (x >= 0) g = 0;else{g = 1;x += 10;}c.s[c.len++] = x;}c.clean();return c;}///乘bign operator * (const bign& b) const{bign c;c.len = len + b.len;for (int i = 0; i < len; i++)for (int j = 0; j < b.len; j++)c.s[i + j] += s[i] * b.s[j];for (int i = 0; i < c.len - 1; i++){c.s[i + 1] += c.s[i] / 10;c.s[i] %= 10;}c.clean();return c;}///除bign operator / (const bign &b) const{bign ret, cur = 0;ret.len = len;for (int i = len - 1; i >= 0; i--){cur = cur * 10;cur.s[0] = s[i];while (cur >= b){cur -= b;ret.s[i]++;}}ret.clean();return ret;}///模、余bign operator % (const bign &b) const{bign c = *this / b;return *this - c * b;}bool operator < (const bign& b) const{if (len != b.len) return len < b.len;for (int i = len - 1; i >= 0; 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 &a) const{return *this > a || *this < a;}bign operator += (const bign &a){*this = *this + a;return *this;}bign operator -= (const bign &a){*this = *this - a;return *this;}bign operator *= (const bign &a){*this = *this * a;return *this;}bign operator /= (const bign &a){*this = *this / a;return *this;}bign operator %= (const bign &a){*this = *this % a;return *this;}} a, b;int main(void){char op;while (~scanf("%s %c %s", numstr, &op, numstr2)){a = numstr, b = numstr2;puts(op == '/' ? (a / b).str() : (a % b).str());}return 0;}