UVA 10023 - Square root(手算开根)

 Square root  

The Problem

You are to determinate X by given Y, from expression 

The Input

The first line is the number of test cases, followed by a blank line.

Each test case of the input contains a positive integer Y (1<=Y<=10¹⁰⁰⁰), with no blanks or leading zeroes in it.
It is guaranteed, that for given Y, X will be always an integer.

Each test case will be separated by a single line.

The Output

For each test case, your program should print X in the same format as Y was given in input.

Print a blank line between the outputs for two consecutive test cases.

Sample Input

17206604678144

Sample Output

2684512

题意:给n 求根号n

思路：高精度， 用手算开根法。

代码:

#include <stdio.h>#include <string.h>#include <math.h>#define max(a,b) (a)>(b)?(a):(b)#define min(a,b) (a)<(b)?(a):(b)const int MAXSIZE = 1005;struct bign {int s[MAXSIZE];bign (){memset(s, 0, sizeof(s));}bign (int number) {*this = number;}bign (const char* number) {*this = number;}    void put();bign mul(int d);void del();    bign operator =  (char *num);bign operator =  (int num);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); }    bign operator + (const bign& c);bign operator * (const bign& c);bign operator - (const bign& c);int  operator / (const bign& c);bign operator / (int k);bign operator % (const bign &c);int  operator % (int k);void operator ++ ();bool operator -- ();};bign 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 bign::operator = (int num) {char str[MAXSIZE];sprintf(str, "%d", num);return *this = str;}bool bign::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;}bign 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 bign::operator * (const bign& c) {bign ans;ans.s[0] = 0; for (int i = 1; i <= c.s[0]; i++){  int g = 0;  for (int j = 1; j <= s[0]; j++){  int x = s[j] * c.s[i] + g + ans.s[i + j - 1];  ans.s[i + j - 1] = x % 10;  g = x / 10;  }  int t = i + s[0] - 1;while (g){  ++t;g += ans.s[t];ans.s[t] = g % 10;g = g / 10;  }  ans.s[0] = max(ans.s[0], t);}  ans.del();return ans;}bign bign::operator - (const bign& c) {bign ans = *this; int i;for (i = 1; i <= c.s[0]; i++) {if (ans.s[i] < c.s[i]) {ans.s[i] += 10;ans.s[i + 1]--;;}ans.s[i] -= c.s[i];}for (i = 1; i <= ans.s[0]; i++) {if (ans.s[i] < 0) {ans.s[i] += 10;ans.s[i + 1]--;}}ans.del();return ans;}int bign::operator / (const bign& c) {int ans = 0;bign d = *this;while (d >= c) {d = d - c;ans++;}return ans;}bign bign::operator / (int k) {bign ans; ans.s[0] = s[0];int num = 0;  for (int i = s[0]; i; i--) {  num = num * 10 + s[i];  ans.s[i] = num / k;  num = num % k;  }  ans.del();return ans;}int bign:: operator % (int k){  int sum = 0;  for (int i = s[0]; i; i--){  sum = sum * 10 + s[i];  sum = sum % k;  }  return sum;  } bign bign::operator % (const bign &c) {bign now = *this;while (now >= c) {now = now - c;now.del();}return now;}void bign::operator ++ () {s[1]++;for (int i = 1; s[i] == 10; i++) {s[i] = 0;s[i + 1]++;s[0] = max(s[0], i + 1);}}bool bign::operator -- () {del();if (s[0] == 1 && s[1] == 0) return false;int i;for (i = 1; s[i] == 0; i++)s[i] = 9;s[i]--;del();return true;}void bign::put() {if (s[0] == 0)printf("0");elsefor (int i = s[0]; i; i--)printf("%d", s[i]);}bign bign::mul(int d) {s[0] += d; int i;for (i = s[0]; i > d; i--)s[i] = s[i - d];for (i = d; i; i--)s[i] = 0;return *this;}void bign::del() {while (s[s[0]] == 0) {s[0]--;if (s[0] == 0) break;}}const int N = 1005;int t, n, len;int num[N];char Num[N];void init() {    memset(num, 0, sizeof(num));    scanf("%s", Num); len = strlen(Num);    int v;    if (len % 2) {num[0] = Num[0] - '0';v = 1;    }    else {num[0] = (Num[0] - '0') * 10 + (Num[1] - '0');v = 2;    }    for (; v < len; v += 2) {num[(v + 1) / 2] = (Num[v] - '0') * 10 + (Num[v + 1] - '0');    }}void solve() {    init(); bign ten = 10, tw = 20, hu = 100;    bign yu; bign sum = 0;    for (int i = 0; i < 10; i ++)if ((i + 1) * (i + 1) > num[0]) {    bign t = i;    sum = sum * ten + t;    yu = (num[0] - i * i);    break;}    bign nu;    for (int i = 1; i < (len + 1) / 2; i ++) {bign t = num[i];nu = yu * hu + t;for (int j = 0; ; j ++) {    bign t = (j + 1);    if ((sum * tw + t) * t > nu) {bign t = j;yu = nu - (sum * tw + t) * t;sum = sum * ten + t;break;    }}    }    sum.put();    printf("\n");    if (t) printf("\n");}int main() {    scanf("%d", &t);     while (t --) {solve();    }    return 0;}