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  • 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] == '')
    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 }
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  • 原文地址:https://www.cnblogs.com/PrayG/p/5732648.html
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