UVa 424 Integer Inquiry （高精度）

UVa 424 Integer Inquiry （高精度）2014-07-31 csdn博客 synapse7424 - Integer Inquiry

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php？option=com_onlinejudge&Itemid=8&category=97&page=show_problem&problem=365

One of the first users of BIT"s new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers.

``This supercomputer is great,"" remarked Chip. ``I only wish Timothy were here to see these results."" （Chip moved to a new apartment, once one became available on the third floor of the Lemon Sky apartments on Third Street.）

Input

The input will consist of at most 100 lines of text, each of which contains a single VeryLongInteger. Each VeryLongInteger will be 100 or fewer characters in length, and will only contain digits （no VeryLongInteger will be negative）.

The final input line will contain a single zero on a line by itself.

Output

Your program should output the sum of the VeryLongIntegers given in the input.

Sample Input

123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0

Sample Output

370370367037037036703703703670

完整代码：

/*0.015s*/#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 105;char numstr[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] = "";}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;}} sum, n;int main（void）{while （n = gets（numstr）, n ！= 0）sum += n;puts（sum.str（））;return 0;}