• 2017 ACM/ICPC 沈阳 F题 Heron and his triangle


    A triangle is a Heron’s triangle if it satisfies that the side lengths of it are consecutive integers t−1, t, t+ 1 and thatits area is an integer. Now, for given n you need to find a Heron’s triangle associated with the smallest t bigger 
    than or equal to n.

    InputThe input contains multiple test cases. The first line of a multiple input is an integer T (1 ≤ T ≤ 30000) followedby T lines. Each line contains an integer N (1 ≤ N ≤ 10^30). 
    OutputFor each test case, output the smallest t in a line. If the Heron’s triangle required does not exist, output -1.Sample Input

    4
    1
    2
    3
    4

    Sample Output

    4
    4
    4
    4
    题解:大数.找规律.a[n]=4*a[n-1]-a[n-2](附上C++大数模板)
      1 #include <bits/stdc++.h>
      2 #define rep(i,a,n) for(int i=a;i<n;++i)
      3 #define per(i,a,n) for(int i=n-1;i>=a;--i)
      4 #define fi first
      5 #define se second
      6 using namespace std;
      7 // base and base_digits must be consistent
      8 constexpr int base = 1000000000;
      9 constexpr int base_digits = 9;
     10 struct bigint{
     11     vector<int> z;
     12     int sign;
     13     bigint() : sign(1) {}
     14     bigint(long long v) { *this = v; }
     15     bigint& operator=(long long v)
     16     {
     17         sign = v < 0 ? -1 : 1;
     18         v*=sign;
     19         z.clear();
     20         for(; v > 0; v = v / base) z.push_back((int)(v % base));
     21         return *this;
     22     }
     23 
     24     bigint(const string& s) { read(s); }
     25 
     26     bigint& operator+=(const bigint& other)
     27     {
     28         if (sign == other.sign)
     29         {
     30             for (int i = 0, carry = 0; i < other.z.size() || carry; ++i)
     31             {
     32                 if(i==z.size()) z.push_back(0);
     33                 z[i] += carry + (i < other.z.size() ? other.z[i] : 0);
     34                 carry = z[i] >= base;
     35                 if(carry) z[i] -= base;
     36             }
     37         }
     38         else if (other != 0 /* prevent infinite loop */)
     39         {
     40             *this -= -other;
     41         }
     42         return *this;
     43     }
     44 
     45     friend bigint operator+(bigint a, const bigint& b)
     46     {
     47         return a += b;
     48     }
     49 
     50     bigint& operator-=(const bigint& other)
     51     {
     52         if (sign == other.sign)
     53         {
     54             if (sign == 1 && *this >= other || sign == -1 && *this <= other)
     55             {
     56                 for (int i = 0, carry = 0; i < other.z.size() || carry; ++i)
     57                 {
     58                     z[i] -= carry + (i < other.z.size() ? other.z[i] : 0);
     59                     carry = z[i] < 0;
     60                     if(carry) z[i] += base;
     61                 }
     62                 trim();
     63             }
     64             else
     65             {
     66                 *this = other - *this;
     67                 this->sign = -this->sign;
     68             }
     69         }
     70         else *this += -other;
     71         return *this;
     72     }
     73 
     74     friend bigint operator - (bigint a,const bigint& b)
     75     {
     76         return a -= b;
     77     }
     78 
     79     bigint& operator*=(int v)
     80     {
     81         if(v<0) sign=-sign,v=-v;
     82         for(int i=0,carry=0;i<z.size() || carry;++i)
     83         {
     84             if(i==z.size()) z.push_back(0);
     85             long long cur = (long long)z[i] * v + carry;
     86             carry = (int)(cur / base);
     87             z[i] = (int)(cur % base);
     88         }
     89         trim();
     90         return *this;
     91     }
     92 
     93     bigint operator*(int v) const
     94     {
     95         return bigint(*this) *= v;
     96     }
     97 
     98     friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1)
     99     {
    100         int norm = base / (b1.z.back() + 1);
    101         bigint a = a1.abs() * norm;
    102         bigint b = b1.abs() * norm;
    103         bigint q, r;
    104         q.z.resize(a.z.size());
    105 
    106         for (int i = (int)a.z.size() - 1; i >= 0; i--)
    107         {
    108             r*=base; r+=a.z[i];
    109             int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
    110             int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
    111             int d = (int)(((long long)s1 * base + s2) / b.z.back());
    112             r -= b * d;
    113             while(r < 0) r+=b,--d;
    114             q.z[i] = d;
    115         }
    116 
    117         q.sign = a1.sign * b1.sign;
    118         r.sign = a1.sign;
    119         q.trim();
    120         r.trim();
    121         return {q, r / norm};
    122     }
    123 
    124     friend bigint sqrt(const bigint& a1)
    125     {
    126         bigint a=a1;
    127         while(a.z.empty()||a.z.size()%2==1) a.z.push_back(0);
    128 
    129         int n = a.z.size();
    130         int firstDigit = (int)::sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
    131         int norm = base / (firstDigit + 1);
    132         a *= norm;
    133         a *= norm;
    134         while(a.z.empty()||a.z.size()%2==1) a.z.push_back(0);
    135 
    136         bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
    137         firstDigit = (int)::sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
    138         int q = firstDigit;
    139         bigint res;
    140         for (int j = n / 2 - 1; j >= 0; j--)
    141         {
    142             for(;;--q)
    143             {
    144                 bigint r1=(r-(res*2*base+q)*q)*base*base+(j>0?(long long)a.z[2*j-1]*base+a.z[2*j-2]:0);
    145                 if(r1>=0) { r=r1; break; }
    146             }
    147             res*=base;res+=q;
    148             if(j>0)
    149             {
    150                 int d1 = res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
    151                 int d2 = res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
    152                 int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()]:0;
    153                 q = (int)(((long long)d1*base*base+(long long)d2*base+d3)/(firstDigit*2));
    154             }
    155         }
    156 
    157         res.trim();
    158         return res / norm;
    159     }
    160 
    161     bigint operator/(const bigint& v) const
    162     {
    163         return divmod(*this, v).first;
    164     }
    165 
    166     bigint operator%(const bigint& v) const
    167     {
    168         return divmod(*this, v).second;
    169     }
    170 
    171     bigint& operator/=(int v)
    172     {
    173         if(v<0) sign=-sign,v=-v;
    174         for (int i = (int)z.size() - 1, rem = 0; i >= 0; --i)
    175         {
    176             long long cur = z[i] + rem * (long long)base;
    177             z[i] = (int)(cur / v);
    178             rem = (int)(cur % v);
    179         }
    180         trim();
    181         return *this;
    182     }
    183 
    184     bigint operator/(int v) const
    185     {
    186         return bigint(*this) /= v;
    187     }
    188 
    189     int operator%(int v) const
    190     {
    191         if(v<0) v=-v;
    192         int m=0;
    193         for(int i=(int)z.size()-1;i>=0;--i) m=(int)((z[i]+m*(long long)base)%v);
    194         return m * sign;
    195     }
    196 
    197     bigint& operator*=(const bigint& v)
    198     {
    199         *this = *this * v;
    200         return *this;
    201     }
    202 
    203     bigint& operator/=(const bigint& v)
    204     {
    205         *this = *this / v;
    206         return *this;
    207     }
    208 
    209     bool operator<(const bigint& v) const
    210     {
    211         if(sign!=v.sign) return sign < v.sign;
    212         if(z.size()!=v.z.size()) return z.size()*sign<v.z.size()*v.sign;
    213         for(int i = (int)z.size() - 1; i >= 0; i--)
    214             if(z[i] != v.z[i])  return z[i] * sign < v.z[i] * sign;
    215         return false;
    216     }
    217 
    218     bool operator>(const bigint& v) const { return v < *this; }
    219     bool operator<=(const bigint& v) const { return !(v < *this); }
    220     bool operator>=(const bigint& v) const { return !(*this < v); }
    221     bool operator==(const bigint& v) const { return !(*this < v) && !(v < *this); }
    222     bool operator!=(const bigint& v) const { return *this < v || v < *this; }
    223 
    224     void trim()
    225     {
    226         while(!z.empty() && z.back() == 0) z.pop_back();
    227         if(z.empty()) sign = 1;
    228     }
    229 
    230     bool isZero() const { return z.empty(); }
    231 
    232     friend bigint operator-(bigint v)
    233     {
    234         if(!v.z.empty()) v.sign = -v.sign;
    235         return v;
    236     }
    237 
    238     bigint abs() const
    239     {
    240         return sign == 1 ? *this : -*this;
    241     }
    242 
    243     long long longValue() const
    244     {
    245         long long res = 0;
    246         for(int i = (int)z.size() - 1; i >= 0; i--) res = res * base + z[i];
    247         return res * sign;
    248     }
    249 
    250     friend bigint gcd(const bigint& a, const bigint& b)
    251     {
    252         return b.isZero() ? a : gcd(b, a % b);
    253     }
    254 
    255     friend bigint lcm(const bigint& a, const bigint& b)
    256     {
    257         return a / gcd(a, b) * b;
    258     }
    259 
    260     void read(const string& s)
    261     {
    262         sign = 1;
    263         z.clear();
    264         int pos = 0;
    265         while(pos < s.size() && (s[pos] == '-' || s[pos] == '+'))
    266         {
    267             if(s[pos] == '-') sign = -sign;
    268             ++pos;
    269         }
    270         for(int i=(int)s.size()-1;i>=pos;i-=base_digits)
    271         {
    272             int x=0;
    273             for(int j=max(pos,i-base_digits+1);j<=i;j++) x=x*10+s[j]-'0';
    274             z.push_back(x);
    275         }
    276         trim();
    277     }
    278 
    279     friend istream& operator>>(istream& stream, bigint& v)
    280     {
    281         string s;
    282         stream >> s;
    283         v.read(s);
    284         return stream;
    285     }
    286 
    287     friend ostream& operator<<(ostream& stream, const bigint& v)
    288     {
    289         if(v.sign == -1) stream << '-';
    290         stream << (v.z.empty() ? 0 : v.z.back());
    291         for(int i = (int)v.z.size() - 2; i >= 0; --i)
    292             stream << setw(base_digits) << setfill('0') << v.z[i];
    293         return stream;
    294     }
    295 
    296     static vector<int> convert_base(const vector<int>& a, int old_digits, int new_digits)
    297     {
    298         vector<long long> p(max(old_digits, new_digits) + 1);
    299         p[0] = 1;
    300         for(int i=1;i<p.size();i++) p[i]=p[i-1]*10;
    301         vector<int> res;
    302         long long cur = 0;
    303         int cur_digits = 0;
    304         for(int v : a)
    305         {
    306             cur += v * p[cur_digits];
    307             cur_digits += old_digits;
    308             while (cur_digits >= new_digits)
    309             {
    310                 res.push_back(int(cur % p[new_digits]));
    311                 cur /= p[new_digits];
    312                 cur_digits -= new_digits;
    313             }
    314         }
    315         res.push_back((int)cur);
    316         while(!res.empty() && res.back()==0)
    317             res.pop_back();
    318         return res;
    319     }
    320 
    321     typedef vector<long long> vll;
    322     static vll karatsubaMultiply(const vll& a, const vll& b)
    323     {
    324         int n=a.size();
    325         vll res(n + n);
    326         if(n <= 32)
    327         {
    328             for (int i = 0; i < n; i++)
    329                 for (int j = 0; j < n; j++)
    330                     res[i + j] += a[i] * b[j];
    331             return res;
    332         }
    333 
    334         int k = n >> 1;
    335         vll a1(a.begin(), a.begin() + k);
    336         vll a2(a.begin() + k, a.end());
    337         vll b1(b.begin(), b.begin() + k);
    338         vll b2(b.begin() + k, b.end());
    339         vll a1b1 = karatsubaMultiply(a1, b1);
    340         vll a2b2 = karatsubaMultiply(a2, b2);
    341         for(int i=0;i<k;i++) a2[i]+=a1[i];
    342         for(int i=0;i<k;i++) b2[i]+=b1[i];
    343 
    344         vll r = karatsubaMultiply(a2, b2);
    345         for(int i=0;i<a1b1.size();i++) r[i]-=a1b1[i];
    346         for(int i=0;i<a2b2.size();i++) r[i]-=a2b2[i];
    347         for(int i=0;i<r.size();i++) res[i+k]+=r[i];
    348         for(int i=0;i<a1b1.size();i++) res[i]+=a1b1[i];
    349         for(int i = 0;i<a2b2.size();i++) res[i+n]+=a2b2[i];
    350         return res;
    351     }
    352 
    353     bigint operator*(const bigint& v) const
    354     {
    355         vector<int> a6=convert_base(this->z,base_digits,6);
    356         vector<int> b6=convert_base(v.z,base_digits,6);
    357         vll a(a6.begin(),a6.end());
    358         vll b(b6.begin(),b6.end());
    359         while(a.size()<b.size()) a.push_back(0);
    360         while(b.size()<a.size()) b.push_back(0);
    361         while(a.size()&(a.size()-1)) a.push_back(0),b.push_back(0);
    362         vll c=karatsubaMultiply(a, b);
    363         bigint res;
    364         res.sign = sign * v.sign;
    365         for (int i = 0, carry = 0; i < c.size(); i++)
    366         {
    367             long long cur = c[i] + carry;
    368             res.z.push_back((int)(cur % 1000000));
    369             carry = (int)(cur / 1000000);
    370         }
    371         res.z = convert_base(res.z, 6, base_digits);
    372         res.trim();
    373         return res;
    374     }
    375 };
    376 
    377 bigint qpow(bigint a,bigint b){
    378     bigint ans=1;
    379     while(b!=0){
    380         if(b%2){
    381             ans= ans*a;
    382         }
    383         b/=2;
    384         a= a*a;
    385     }
    386     return ans;
    387 
    388 }
    389 
    390 
    391 struct Matrix
    392 {
    393     bigint a[2][2];
    394     Matrix()
    395     {
    396         rep(i,0,2){
    397             rep(j,0,2){
    398                 a[i][j]=0;
    399             }
    400         }
    401     }
    402     Matrix operator * (const Matrix y)
    403     {
    404         Matrix ans;
    405         for(int i = 0; i < 2; i++)
    406             for(int j = 0; j < 2; j++)
    407                 for(int k = 0; k < 2; k++)
    408                     ans.a[i][j] = ans.a[i][j] + (a[i][k]*y.a[k][j]);
    409         return ans;
    410     }
    411     Matrix operator = (const Matrix y)
    412     {
    413         for(int i=0;i<2;i++)
    414             for(int j=0;j<2;j++)
    415                 a[i][j]=y.a[i][j];
    416     }
    417     Matrix operator *= (const Matrix y)
    418     {
    419         Matrix ans;
    420         for(int i=0;i<2;i++)
    421             for(int j=0;j<2;j++)
    422                 for(int k=0;k<2;k++)
    423                     ans.a[i][j] = ans.a[i][j] + (a[i][k]*y.a[k][j]);
    424 
    425         for(int i=0;i<2;i++)
    426             for(int j=0;j<2;j++)
    427                 a[i][j]=ans.a[i][j];
    428     }
    429 };
    430 
    431 Matrix qpow(bigint x)
    432 {
    433     Matrix ans;
    434     ans.a[0][0]=ans.a[1][1]=1; //单位矩阵
    435     Matrix mul;
    436     mul.a[0][0]=4;
    437     mul.a[0][1]=-1;
    438     mul.a[1][0]=1;
    439     mul.a[1][1]=0;
    440     while(x!=0)
    441     {
    442         if(x%2!=0)
    443             ans = ans*mul;
    444         mul = mul* mul;
    445         x/=2;
    446     }
    447     return ans;
    448 }
    449 bigint ans[1005];
    450 void solve(){
    451 
    452 
    453     ans[0]=(bigint)4;
    454     ans[1]=(bigint)14;
    455     ans[2]=(bigint)52;
    456     rep(i,2,200){
    457         ans[i]=(bigint)4*ans[i-1]-ans[i-2];
    458 //        cout<<ans[i]<<endl;
    459     }
    460 
    461 }
    462 int main()
    463 {
    464     solve();
    465     int t;cin>>t;
    466     while(t--){
    467         bigint n;
    468         cin>>n;
    469         rep(i,0,200){
    470             if(ans[i]>=n){
    471                 cout<<ans[i]<<endl;
    472                 break;
    473             }
    474         }
    475 
    476 
    477     }
    478 }
    View Code

      

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  • 原文地址:https://www.cnblogs.com/csushl/p/9784961.html
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