Sgu 111

111. Very simple problem
time limit per test: 0.25 sec. memory limit per test: 4096 KB
You are given natural number X. Find such maximum integer number that it square is not greater than X.
Input
Input file contains number X (1≤X≤10¹⁰⁰⁰).
Output
Write answer in output file.
Sample Input
Sample Output
  1 #pragma comment(linker, "/STACK:1024000000,1024000000")
  2 #include <stdio.h>
  3 #include <string.h>
  4 #include <iostream>
  5 #include <algorithm>
  6 #include <vector>
  7 #include <queue>
  8 #include <set>
  9 #include <map>
 10 #include <string>
 11 #include <math.h>
 12 #include <stdlib.h>
 13 #include <time.h>
 14 using namespace std;
 15 /*
 16 * 完全大数模板
 17 * 输出cin>>a
 18 * 输出a.print();
 19 * 注意这个输入不能自动去掉前导0的，可以先读入到char数组，去掉前导0，再用构造函数。
 20 */
 21 #define MAXN 9999
 22 #define MAXSIZE 1010
 23 #define DLEN 4
 24 
 25 class BigNum
 26 {
 27 private:
 28     int a[MAXN];  //可以控制大数的位数
 29     int len;
 30 public:
 31     BigNum(){ len = 1; memset(a, 0, sizeof(a)); }  //构造函数
 32     BigNum(const int);     //将一个int类型的变量转化成大数
 33     BigNum(const char*);   //将一个字符串类型的变量转化为大数
 34     BigNum(const BigNum &); //拷贝构造函数
 35     BigNum &operator=(const BigNum &); //重载赋值运算符，大数之间进行赋值运算
 36     friend istream& operator>>(istream&, BigNum&); //重载输入运算符
 37     friend ostream& operator<<(ostream&, BigNum&); //重载输出运算符
 38 
 39     BigNum operator+(const BigNum &)const;  //重载加法运算符，两个大数之间的相加运算
 40     BigNum operator-(const BigNum &)const;  //重载减法运算符，两个大数之间的相减运算
 41     BigNum operator*(const BigNum &)const;  //重载乘法运算符，两个大数之间的相乘运算
 42     BigNum operator/(const int &)const;     //重载除法运算符，大数对一个整数进行相除运算
 43 
 44     BigNum operator^(const int &)const;     //大数的n次方运算
 45     int operator%(const int &)const;        //大数对一个int类型的变量进行取模运算
 46     bool operator>(const BigNum &T)const;   //大数和另一个大数的大小比较
 47     bool operator>(const int &t)const;      //大数和一个int类型的变量的大小比较
 48 
 49     void print();        //输出大数
 50 };
 51 BigNum::BigNum(const int b)   //将一个int类型的变量转化为大数
 52 {
 53     int c, d = b;
 54     len = 0;
 55     memset(a, 0, sizeof(a));
 56     while (d>MAXN)
 57     {
 58         c = d - (d / (MAXN + 1))*(MAXN + 1);
 59         d = d / (MAXN + 1);
 60         a[len++] = c;
 61     }
 62     a[len++] = d;
 63 }
 64 BigNum::BigNum(const char *s)  //将一个字符串类型的变量转化为大数
 65 {
 66     int t, k, index, L, i;
 67     memset(a, 0, sizeof(a));
 68     L = strlen(s);
 69     len = L / DLEN;
 70     if (L%DLEN)len++;
 71     index = 0;
 72     for (i = L - 1; i >= 0; i -= DLEN)
 73     {
 74         t = 0;
 75         k = i - DLEN + 1;
 76         if (k<0)k = 0;
 77         for (int j = k; j <= i; j++)
 78             t = t * 10 + s[j] - '0';
 79         a[index++] = t;
 80     }
 81 }
 82 BigNum::BigNum(const BigNum &T) :len(T.len)  //拷贝构造函数
 83 {
 84     int i;
 85     memset(a, 0, sizeof(a));
 86     for (i = 0; i<len; i++)
 87         a[i] = T.a[i];
 88 }
 89 BigNum & BigNum::operator=(const BigNum &n)  //重载赋值运算符，大数之间赋值运算
 90 {
 91     int i;
 92     len = n.len;
 93     memset(a, 0, sizeof(a));
 94     for (i = 0; i<len; i++)
 95         a[i] = n.a[i];
 96     return *this;
 97 }
 98 istream& operator>>(istream &in, BigNum &b)
 99 {
100     char ch[MAXSIZE * 4];
101     int i = -1;
102     in >> ch;
103     int L = strlen(ch);
104     int count = 0, sum = 0;
105     for (i = L - 1; i >= 0;)
106     {
107         sum = 0;
108         int t = 1;
109         for (int j = 0; j<4 && i >= 0; j++, i--, t *= 10)
110         {
111             sum += (ch[i] - '0')*t;
112         }
113         b.a[count] = sum;
114         count++;
115     }
116     b.len = count++;
117     return in;
118 }
119 ostream& operator<<(ostream& out, BigNum& b)  //重载输出运算符
120 {
121     int i;
122     cout << b.a[b.len - 1];
123     for (i = b.len - 2; i >= 0; i--)
124     {
125         printf("%04d", b.a[i]);
126     }
127     return out;
128 }
129 BigNum BigNum::operator+(const BigNum &T)const   //两个大数之间的相加运算
130 {
131     BigNum t(*this);
132     int i, big;
133     big = T.len>len ? T.len : len;
134     for (i = 0; i<big; i++)
135     {
136         t.a[i] += T.a[i];
137         if (t.a[i]>MAXN)
138         {
139             t.a[i + 1]++;
140             t.a[i] -= MAXN + 1;
141         }
142     }
143     if (t.a[big] != 0)
144         t.len = big + 1;
145     else t.len = big;
146     return t;
147 }
148 BigNum BigNum::operator-(const BigNum &T)const  //两个大数之间的相减运算
149 {
150     int i, j, big;
151     bool flag;
152     BigNum t1, t2;
153     if (*this>T)
154     {
155         t1 = *this;
156         t2 = T;
157         flag = 0;
158     }
159     else
160     {
161         t1 = T;
162         t2 = *this;
163         flag = 1;
164     }
165     big = t1.len;
166     for (i = 0; i<big; i++)
167     {
168         if (t1.a[i]<t2.a[i])
169         {
170             j = i + 1;
171             while (t1.a[j] == 0)
172                 j++;
173             t1.a[j--]--;
174             while (j>i)
175                 t1.a[j--] += MAXN;
176             t1.a[i] += MAXN + 1 - t2.a[i];
177         }
178         else t1.a[i] -= t2.a[i];
179     }
180     t1.len = big;
181     while (t1.a[len - 1] == 0 && t1.len>1)
182     {
183         t1.len--;
184         big--;
185     }
186     if (flag)
187         t1.a[big - 1] = 0 - t1.a[big - 1];
188     return t1;
189 }
190 BigNum BigNum::operator*(const BigNum &T)const  //两个大数之间的相乘
191 {
192     BigNum ret;
193     int i, j, up;
194     int temp, temp1;
195     for (i = 0; i<len; i++)
196     {
197         up = 0;
198         for (j = 0; j<T.len; j++)
199         {
200             temp = a[i] * T.a[j] + ret.a[i + j] + up;
201             if (temp>MAXN)
202             {
203                 temp1 = temp - temp / (MAXN + 1)*(MAXN + 1);
204                 up = temp / (MAXN + 1);
205                 ret.a[i + j] = temp1;
206             }
207             else
208             {
209                 up = 0;
210                 ret.a[i + j] = temp;
211             }
212         }
213         if (up != 0)
214             ret.a[i + j] = up;
215     }
216     ret.len = i + j;
217     while (ret.a[ret.len - 1] == 0 && ret.len>1)ret.len--;
218     return ret;
219 }
220 BigNum BigNum::operator/(const int &b)const  //大数对一个整数进行相除运算
221 {
222     BigNum ret;
223     int i, down = 0;
224     for (i = len - 1; i >= 0; i--)
225     {
226         ret.a[i] = (a[i] + down*(MAXN + 1)) / b;
227         down = a[i] + down*(MAXN + 1) - ret.a[i] * b;
228     }
229     ret.len = len;
230     while (ret.a[ret.len - 1] == 0 && ret.len>1)
231         ret.len--;
232     return ret;
233 }
234 int BigNum::operator%(const int &b)const   //大数对一个 int类型的变量进行取模
235 {
236     int i, d = 0;
237     for (i = len - 1; i >= 0; i--)
238         d = ((d*(MAXN + 1)) % b + a[i]) % b;
239     return d;
240 }
241 BigNum BigNum::operator^(const int &n)const  //大数的n次方运算
242 {
243     BigNum t, ret(1);
244     int i;
245     if (n<0)exit(-1);
246     if (n == 0)return 1;
247     if (n == 1)return *this;
248     int m = n;
249     while (m>1)
250     {
251         t = *this;
252         for (i = 1; (i << 1) <= m; i <<= 1)
253             t = t*t;
254         m -= i;
255         ret = ret*t;
256         if (m == 1)ret = ret*(*this);
257     }
258     return ret;
259 }
260 bool BigNum::operator>(const BigNum &T)const    //大数和另一个大数的大小比较
261 {
262     int ln;
263     if (len>T.len)return true;
264     else if (len == T.len)
265     {
266         ln = len - 1;
267         while (a[ln] == T.a[ln] && ln >= 0)
268             ln--;
269         if (ln >= 0 && a[ln]>T.a[ln])
270             return true;
271         else
272             return false;
273     }
274     else
275         return false;
276 }
277 bool BigNum::operator>(const int &t)const  //大数和一个int类型的变量的大小比较
278 {
279     BigNum b(t);
280     return *this>b;
281 }
282 void BigNum::print()   //输出大数
283 {
284     int i;
285     printf("%d", a[len - 1]);
286     for (i = len - 2; i >= 0; i--)
287         printf("%04d", a[i]);
288     printf("\n");
289 }
290 BigNum x;
291 void fun(){
292     BigNum l = 0, r = 1, mid;
293     for (int i = 0; i < 1000; i++)r =r* 10;
294     while (r > l){
295         mid = (l + r) / 2;
296         if (mid*mid>x)r = mid;
297         else l = mid + 1;
298     }
299     cout << l - 1 << endl;
300 }
301 int main(){
302     cin >> x;
303     fun();
304     return 0;
305 }
View Code
posted @ 2014-01-29 18:40 HaibaraAi 阅读(160) 评论(0) 收藏举报
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