gym 100735I

Description

standard input/output
Statements

You are given three numbers. Is there a way to replace variables A, B and C with these numbers so the equality A + B = C is correct?

Input

There are three numbers X1, X2 and X3 (1 ≤ Xi ≤ 10100), each on a separate line of input.

Output

Output either "YES", if there is a way to substitute variables A, B and C with given numbers so the equality is correct, or "NO" otherwise.

Sample Input

Input
1
2
3
Output
YES
Input
1
2
4
Output
YES
Input
1
3
5
Output
NO
题意： 给你3 个数，让你判断其中任意两个数的和是否可以组成这三个树种的一个，这些数都可以重复使用。
分析： 高精度
  1 #include<string>
  2 #include<iostream>
  3 #include<iosfwd>
  4 #include<cmath>
  5 #include<cstring>
  6 #include<stdlib.h>
  7 #include<stdio.h>
  8 #include<cstring>
  9 #define MAX_L 2005 //最大长度，可以修改
 10 using namespace std;
 11 
 12 class bign
 13 {
 14 public:
 15     int len, s[MAX_L];//数的长度，记录数组
 16 //构造函数
 17     bign();
 18     bign(const char*);
 19     bign(int);
 20     bool sign;//符号 1正数 0负数
 21     string toStr() const;//转化为字符串，主要是便于输出
 22     friend istream& operator>>(istream &,bign &);//重载输入流
 23     friend ostream& operator<<(ostream &,bign &);//重载输出流
 24 //重载复制
 25     bign operator=(const char*);
 26     bign operator=(int);
 27     bign operator=(const string);
 28 //重载各种比较
 29     bool operator>(const bign &) const;
 30     bool operator>=(const bign &) const;
 31     bool operator<(const bign &) const;
 32     bool operator<=(const bign &) const;
 33     bool operator==(const bign &) const;
 34     bool operator!=(const bign &) const;
 35 //重载四则运算
 36     bign operator+(const bign &) const;
 37     bign operator++();
 38     bign operator++(int);
 39     bign operator+=(const bign&);
 40     bign operator-(const bign &) const;
 41     bign operator--();
 42     bign operator--(int);
 43     bign operator-=(const bign&);
 44     bign operator*(const bign &)const;
 45     bign operator*(const int num)const;
 46     bign operator*=(const bign&);
 47     bign operator/(const bign&)const;
 48     bign operator/=(const bign&);
 49 //四则运算的衍生运算
 50     bign operator%(const bign&)const;//取模（余数）
 51     bign factorial()const;//阶乘
 52     bign Sqrt()const;//整数开根（向下取整）
 53     bign pow(const bign&)const;//次方
 54 //一些乱乱的函数
 55     void clean();
 56     ~bign();
 57 };
 58 #define max(a,b) a>b ? a : b
 59 #define min(a,b) a<b ? a : b
 60 
 61 bign::bign()
 62 {
 63     memset(s, 0, sizeof(s));
 64     len = 1;
 65     sign = 1;
 66 }
 67 
 68 bign::bign(const char *num)
 69 {
 70     *this = num;
 71 }
 72 
 73 bign::bign(int num)
 74 {
 75     *this = num;
 76 }
 77 
 78 string bign::toStr() const
 79 {
 80     string res;
 81     res = "";
 82     for (int i = 0; i < len; i++)
 83         res = (char)(s[i] + '0') + res;
 84     if (res == "")
 85         res = "0";
 86     if (!sign&&res != "0")
 87         res = "-" + res;
 88     return res;
 89 }
 90 
 91 istream &operator>>(istream &in, bign &num)
 92 {
 93     string str;
 94     in>>str;
 95     num=str;
 96     return in;
 97 }
 98 
 99 ostream &operator<<(ostream &out, bign &num)
100 {
101     out<<num.toStr();
102     return out;
103 }
104 
105 bign bign::operator=(const char *num)
106 {
107     memset(s, 0, sizeof(s));
108     char a[MAX_L] = "";
109     if (num[0] != '-')
110         strcpy(a, num);
111     else
112         for (int i = 1; i < strlen(num); i++)
113             a[i - 1] = num[i];
114     sign = !(num[0] == '-');
115     len = strlen(a);
116     for (int i = 0; i < strlen(a); i++)
117         s[i] = a[len - i - 1] - 48;
118     return *this;
119 }
120 
121 bign bign::operator=(int num)
122 {
123     char temp[MAX_L];
124     sprintf(temp, "%d", num);
125     *this = temp;
126     return *this;
127 }
128 
129 bign bign::operator=(const string num)
130 {
131     const char *tmp;
132     tmp = num.c_str();
133     *this = tmp;
134     return *this;
135 }
136 
137 bool bign::operator<(const bign &num) const
138 {
139     if (sign^num.sign)
140         return num.sign;
141     if (len != num.len)
142         return len < num.len;
143     for (int i = len - 1; i >= 0; i--)
144         if (s[i] != num.s[i])
145             return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i]));
146     return !sign;
147 }
148 
149 bool bign::operator>(const bign&num)const
150 {
151     return num < *this;
152 }
153 
154 bool bign::operator<=(const bign&num)const
155 {
156     return !(*this>num);
157 }
158 
159 bool bign::operator>=(const bign&num)const
160 {
161     return !(*this<num);
162 }
163 
164 bool bign::operator!=(const bign&num)const
165 {
166     return *this > num || *this < num;
167 }
168 
169 bool bign::operator==(const bign&num)const
170 {
171     return !(num != *this);
172 }
173 
174 bign bign::operator+(const bign &num) const
175 {
176     if (sign^num.sign)
177     {
178         bign tmp = sign ? num : *this;
179         tmp.sign = 1;
180         return sign ? *this - tmp : num - tmp;
181     }
182     bign result;
183     result.len = 0;
184     int temp = 0;
185     for (int i = 0; temp || i < (max(len, num.len)); i++)
186     {
187         int t = s[i] + num.s[i] + temp;
188         result.s[result.len++] = t % 10;
189         temp = t / 10;
190     }
191     result.sign = sign;
192     return result;
193 }
194 
195 bign bign::operator++()
196 {
197     *this = *this + 1;
198     return *this;
199 }
200 
201 bign bign::operator++(int)
202 {
203     bign old = *this;
204     ++(*this);
205     return old;
206 }
207 
208 bign bign::operator+=(const bign &num)
209 {
210     *this = *this + num;
211     return *this;
212 }
213 
214 bign bign::operator-(const bign &num) const
215 {
216     bign b=num,a=*this;
217     if (!num.sign && !sign)
218     {
219         b.sign=1;
220         a.sign=1;
221         return b-a;
222     }
223     if (!b.sign)
224     {
225         b.sign=1;
226         return a+b;
227     }
228     if (!a.sign)
229     {
230         a.sign=1;
231         b=bign(0)-(a+b);
232         return b;
233     }
234     if (a<b)
235     {
236         bign c=(b-a);
237         c.sign=false;
238         return c;
239     }
240     bign result;
241     result.len = 0;
242     for (int i = 0, g = 0; i < a.len; i++)
243     {
244         int x = a.s[i] - g;
245         if (i < b.len) x -= b.s[i];
246         if (x >= 0) g = 0;
247         else
248         {
249             g = 1;
250             x += 10;
251         }
252         result.s[result.len++] = x;
253     }
254     result.clean();
255     return result;
256 }
257 
258 bign bign::operator * (const bign &num)const
259 {
260     bign result;
261     result.len = len + num.len;
262 
263     for (int i = 0; i < len; i++)
264         for (int j = 0; j < num.len; j++)
265             result.s[i + j] += s[i] * num.s[j];
266 
267     for (int i = 0; i < result.len; i++)
268     {
269         result.s[i + 1] += result.s[i] / 10;
270         result.s[i] %= 10;
271     }
272     result.clean();
273     result.sign = !(sign^num.sign);
274     return result;
275 }
276 
277 bign bign::operator*(const int num)const
278 {
279     bign x = num;
280     bign z = *this;
281     return x*z;
282 }
283 bign bign::operator*=(const bign&num)
284 {
285     *this = *this * num;
286     return *this;
287 }
288 
289 bign bign::operator /(const bign&num)const
290 {
291     bign ans;
292     ans.len = len - num.len + 1;
293     if (ans.len < 0)
294     {
295         ans.len = 1;
296         return ans;
297     }
298 
299     bign divisor = *this, divid = num;
300     divisor.sign = divid.sign = 1;
301     int k = ans.len - 1;
302     int j = len - 1;
303     while (k >= 0)
304     {
305         while (divisor.s[j] == 0) j--;
306         if (k > j) k = j;
307         char z[MAX_L];
308         memset(z, 0, sizeof(z));
309         for (int i = j; i >= k; i--)
310             z[j - i] = divisor.s[i] + '0';
311         bign dividend = z;
312         if (dividend < divid) { k--; continue; }
313         int key = 0;
314         while (divid*key <= dividend) key++;
315         key--;
316         ans.s[k] = key;
317         bign temp = divid*key;
318         for (int i = 0; i < k; i++)
319             temp = temp * 10;
320         divisor = divisor - temp;
321         k--;
322     }
323     ans.clean();
324     ans.sign = !(sign^num.sign);
325     return ans;
326 }
327 
328 bign bign::operator/=(const bign&num)
329 {
330     *this = *this / num;
331     return *this;
332 }
333 
334 bign bign::operator%(const bign& num)const
335 {
336     bign a = *this, b = num;
337     a.sign = b.sign = 1;
338     bign result, temp = a / b*b;
339     result = a - temp;
340     result.sign = sign;
341     return result;
342 }
343 
344 bign bign::pow(const bign& num)const
345 {
346     bign result = 1;
347     for (bign i = 0; i < num; i++)
348         result = result*(*this);
349     return result;
350 }
351 
352 bign bign::factorial()const
353 {
354     bign result = 1;
355     for (bign i = 1; i <= *this; i++)
356         result *= i;
357     return result;
358 }
359 
360 void bign::clean()
361 {
362     if (len == 0) len++;
363     while (len > 1 && s[len - 1] == '\0')
364         len--;
365 }
366 
367 bign bign::Sqrt()const
368 {
369     if(*this<0)return -1;
370     if(*this<=1)return *this;
371     bign l=0,r=*this,mid;
372     while(r-l>1)
373     {
374         mid=(l+r)/2;
375         if(mid*mid>*this)
376             r=mid;
377         else
378             l=mid;
379     }
380     return l;
381 }
382 
383 bign::~bign()
384 {
385 }
386 
387 bign a,b,c;
388 
389 int main()
390 {
391     cin >> a >> b >> c;
392     bool flag = false;
393     if(a + b == c) flag = true;
394     if(a + c == b) flag = true;
395     if(a + a == b) flag = true;
396     if(a + a == c) flag = true;
397     if(b + b == a) flag = true;
398     if(b + b == c) flag = true;
399     if(b + c == a) flag = true;
400     if(c + c == a) flag = true;
401     if(c + c == b) flag = true;
402     if(flag) cout << "YES" << endl;
403     else cout << "NO" << endl;
404     return 0;
405 }

 

posted @ 2016-08-03 13:55  PrayG  阅读(270)  评论(0)  编辑  收藏  举报