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)
 #include <bits/stdc++.h>
#define rep(i,a,n) for(int i=a;i<n;++i)
#define per(i,a,n) for(int i=n-1;i>=a;--i)
#define fi first
#define se second
using namespace std;
// base and base_digits must be consistent
constexpr int base = ;
constexpr int base_digits = ;
struct bigint{
vector<int> z;
int sign;
bigint() : sign() {}
bigint(long long v) { *this = v; }
bigint& operator=(long long v)
{
sign = v < ? - : ;
v*=sign;
z.clear();
for(; v > ; v = v / base) z.push_back((int)(v % base));
return *this;
} bigint(const string& s) { read(s); } bigint& operator+=(const bigint& other)
{
if (sign == other.sign)
{
for (int i = , carry = ; i < other.z.size() || carry; ++i)
{
if(i==z.size()) z.push_back();
z[i] += carry + (i < other.z.size() ? other.z[i] : );
carry = z[i] >= base;
if(carry) z[i] -= base;
}
}
else if (other != /* prevent infinite loop */)
{
*this -= -other;
}
return *this;
} friend bigint operator+(bigint a, const bigint& b)
{
return a += b;
} bigint& operator-=(const bigint& other)
{
if (sign == other.sign)
{
if (sign == && *this >= other || sign == - && *this <= other)
{
for (int i = , carry = ; i < other.z.size() || carry; ++i)
{
z[i] -= carry + (i < other.z.size() ? other.z[i] : );
carry = z[i] < ;
if(carry) z[i] += base;
}
trim();
}
else
{
*this = other - *this;
this->sign = -this->sign;
}
}
else *this += -other;
return *this;
} friend bigint operator - (bigint a,const bigint& b)
{
return a -= b;
} bigint& operator*=(int v)
{
if(v<) sign=-sign,v=-v;
for(int i=,carry=;i<z.size() || carry;++i)
{
if(i==z.size()) z.push_back();
long long cur = (long long)z[i] * v + carry;
carry = (int)(cur / base);
z[i] = (int)(cur % base);
}
trim();
return *this;
} bigint operator*(int v) const
{
return bigint(*this) *= v;
} friend pair<bigint, bigint> divmod(const bigint& a1, const bigint& b1)
{
int norm = base / (b1.z.back() + );
bigint a = a1.abs() * norm;
bigint b = b1.abs() * norm;
bigint q, r;
q.z.resize(a.z.size()); for (int i = (int)a.z.size() - ; i >= ; i--)
{
r*=base; r+=a.z[i];
int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : ;
int s2 = b.z.size() - < r.z.size() ? r.z[b.z.size() - ] : ;
int d = (int)(((long long)s1 * base + s2) / b.z.back());
r -= b * d;
while(r < ) r+=b,--d;
q.z[i] = d;
} q.sign = a1.sign * b1.sign;
r.sign = a1.sign;
q.trim();
r.trim();
return {q, r / norm};
} friend bigint sqrt(const bigint& a1)
{
bigint a=a1;
while(a.z.empty()||a.z.size()%==) a.z.push_back(); int n = a.z.size();
int firstDigit = (int)::sqrt((double)a.z[n - ] * base + a.z[n - ]);
int norm = base / (firstDigit + );
a *= norm;
a *= norm;
while(a.z.empty()||a.z.size()%==) a.z.push_back(); bigint r = (long long)a.z[n - ] * base + a.z[n - ];
firstDigit = (int)::sqrt((double)a.z[n - ] * base + a.z[n - ]);
int q = firstDigit;
bigint res;
for (int j = n / - ; j >= ; j--)
{
for(;;--q)
{
bigint r1=(r-(res**base+q)*q)*base*base+(j>?(long long)a.z[*j-]*base+a.z[*j-]:);
if(r1>=) { r=r1; break; }
}
res*=base;res+=q;
if(j>)
{
int d1 = res.z.size() + < r.z.size() ? r.z[res.z.size() + ] : ;
int d2 = res.z.size() + < r.z.size() ? r.z[res.z.size() + ] : ;
int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()]:;
q = (int)(((long long)d1*base*base+(long long)d2*base+d3)/(firstDigit*));
}
} res.trim();
return res / norm;
} bigint operator/(const bigint& v) const
{
return divmod(*this, v).first;
} bigint operator%(const bigint& v) const
{
return divmod(*this, v).second;
} bigint& operator/=(int v)
{
if(v<) sign=-sign,v=-v;
for (int i = (int)z.size() - , rem = ; i >= ; --i)
{
long long cur = z[i] + rem * (long long)base;
z[i] = (int)(cur / v);
rem = (int)(cur % v);
}
trim();
return *this;
} bigint operator/(int v) const
{
return bigint(*this) /= v;
} int operator%(int v) const
{
if(v<) v=-v;
int m=;
for(int i=(int)z.size()-;i>=;--i) m=(int)((z[i]+m*(long long)base)%v);
return m * sign;
} bigint& operator*=(const bigint& v)
{
*this = *this * v;
return *this;
} bigint& operator/=(const bigint& v)
{
*this = *this / v;
return *this;
} bool operator<(const bigint& v) const
{
if(sign!=v.sign) return sign < v.sign;
if(z.size()!=v.z.size()) return z.size()*sign<v.z.size()*v.sign;
for(int i = (int)z.size() - ; i >= ; i--)
if(z[i] != v.z[i]) return z[i] * sign < v.z[i] * sign;
return false;
} bool operator>(const bigint& v) const { return v < *this; }
bool operator<=(const bigint& v) const { return !(v < *this); }
bool operator>=(const bigint& v) const { return !(*this < v); }
bool operator==(const bigint& v) const { return !(*this < v) && !(v < *this); }
bool operator!=(const bigint& v) const { return *this < v || v < *this; } void trim()
{
while(!z.empty() && z.back() == ) z.pop_back();
if(z.empty()) sign = ;
} bool isZero() const { return z.empty(); } friend bigint operator-(bigint v)
{
if(!v.z.empty()) v.sign = -v.sign;
return v;
} bigint abs() const
{
return sign == ? *this : -*this;
} long long longValue() const
{
long long res = ;
for(int i = (int)z.size() - ; i >= ; i--) res = res * base + z[i];
return res * sign;
} friend bigint gcd(const bigint& a, const bigint& b)
{
return b.isZero() ? a : gcd(b, a % b);
} friend bigint lcm(const bigint& a, const bigint& b)
{
return a / gcd(a, b) * b;
} void read(const string& s)
{
sign = ;
z.clear();
int pos = ;
while(pos < s.size() && (s[pos] == '-' || s[pos] == '+'))
{
if(s[pos] == '-') sign = -sign;
++pos;
}
for(int i=(int)s.size()-;i>=pos;i-=base_digits)
{
int x=;
for(int j=max(pos,i-base_digits+);j<=i;j++) x=x*+s[j]-'';
z.push_back(x);
}
trim();
} friend istream& operator>>(istream& stream, bigint& v)
{
string s;
stream >> s;
v.read(s);
return stream;
} friend ostream& operator<<(ostream& stream, const bigint& v)
{
if(v.sign == -) stream << '-';
stream << (v.z.empty() ? : v.z.back());
for(int i = (int)v.z.size() - ; i >= ; --i)
stream << setw(base_digits) << setfill('') << v.z[i];
return stream;
} static vector<int> convert_base(const vector<int>& a, int old_digits, int new_digits)
{
vector<long long> p(max(old_digits, new_digits) + );
p[] = ;
for(int i=;i<p.size();i++) p[i]=p[i-]*;
vector<int> res;
long long cur = ;
int cur_digits = ;
for(int v : a)
{
cur += v * p[cur_digits];
cur_digits += old_digits;
while (cur_digits >= new_digits)
{
res.push_back(int(cur % p[new_digits]));
cur /= p[new_digits];
cur_digits -= new_digits;
}
}
res.push_back((int)cur);
while(!res.empty() && res.back()==)
res.pop_back();
return res;
} typedef vector<long long> vll;
static vll karatsubaMultiply(const vll& a, const vll& b)
{
int n=a.size();
vll res(n + n);
if(n <= )
{
for (int i = ; i < n; i++)
for (int j = ; j < n; j++)
res[i + j] += a[i] * b[j];
return res;
} int k = n >> ;
vll a1(a.begin(), a.begin() + k);
vll a2(a.begin() + k, a.end());
vll b1(b.begin(), b.begin() + k);
vll b2(b.begin() + k, b.end());
vll a1b1 = karatsubaMultiply(a1, b1);
vll a2b2 = karatsubaMultiply(a2, b2);
for(int i=;i<k;i++) a2[i]+=a1[i];
for(int i=;i<k;i++) b2[i]+=b1[i]; vll r = karatsubaMultiply(a2, b2);
for(int i=;i<a1b1.size();i++) r[i]-=a1b1[i];
for(int i=;i<a2b2.size();i++) r[i]-=a2b2[i];
for(int i=;i<r.size();i++) res[i+k]+=r[i];
for(int i=;i<a1b1.size();i++) res[i]+=a1b1[i];
for(int i = ;i<a2b2.size();i++) res[i+n]+=a2b2[i];
return res;
} bigint operator*(const bigint& v) const
{
vector<int> a6=convert_base(this->z,base_digits,);
vector<int> b6=convert_base(v.z,base_digits,);
vll a(a6.begin(),a6.end());
vll b(b6.begin(),b6.end());
while(a.size()<b.size()) a.push_back();
while(b.size()<a.size()) b.push_back();
while(a.size()&(a.size()-)) a.push_back(),b.push_back();
vll c=karatsubaMultiply(a, b);
bigint res;
res.sign = sign * v.sign;
for (int i = , carry = ; i < c.size(); i++)
{
long long cur = c[i] + carry;
res.z.push_back((int)(cur % ));
carry = (int)(cur / );
}
res.z = convert_base(res.z, , base_digits);
res.trim();
return res;
}
}; bigint qpow(bigint a,bigint b){
bigint ans=;
while(b!=){
if(b%){
ans= ans*a;
}
b/=;
a= a*a;
}
return ans; } struct Matrix
{
bigint a[][];
Matrix()
{
rep(i,,){
rep(j,,){
a[i][j]=;
}
}
}
Matrix operator * (const Matrix y)
{
Matrix ans;
for(int i = ; i < ; i++)
for(int j = ; j < ; j++)
for(int k = ; k < ; k++)
ans.a[i][j] = ans.a[i][j] + (a[i][k]*y.a[k][j]);
return ans;
}
Matrix operator = (const Matrix y)
{
for(int i=;i<;i++)
for(int j=;j<;j++)
a[i][j]=y.a[i][j];
}
Matrix operator *= (const Matrix y)
{
Matrix ans;
for(int i=;i<;i++)
for(int j=;j<;j++)
for(int k=;k<;k++)
ans.a[i][j] = ans.a[i][j] + (a[i][k]*y.a[k][j]); for(int i=;i<;i++)
for(int j=;j<;j++)
a[i][j]=ans.a[i][j];
}
}; Matrix qpow(bigint x)
{
Matrix ans;
ans.a[][]=ans.a[][]=; //单位矩阵
Matrix mul;
mul.a[][]=;
mul.a[][]=-;
mul.a[][]=;
mul.a[][]=;
while(x!=)
{
if(x%!=)
ans = ans*mul;
mul = mul* mul;
x/=;
}
return ans;
}
bigint ans[];
void solve(){ ans[]=(bigint);
ans[]=(bigint);
ans[]=(bigint);
rep(i,,){
ans[i]=(bigint)*ans[i-]-ans[i-];
// cout<<ans[i]<<endl;
} }
int main()
{
solve();
int t;cin>>t;
while(t--){
bigint n;
cin>>n;
rep(i,,){
if(ans[i]>=n){
cout<<ans[i]<<endl;
break;
}
} }
}

  

最新文章

  1. GO语言总结(5)——类型转换和类型断言
  2. C#Color对象的使用介绍及颜色对照表
  3. C# WM_NCMOUSELEAVE 消息触发
  4. Java中怎样由枚举常量的ordinal值获得枚举常量对象
  5. 如何在mysql中退出当前窗口界面而不关闭窗口的方法
  6. Xcode7插件开发:从开发到拉到恶魔岛
  7. 软件测试学习日志———— round 2 Junit+intellj idea 安装及简单的测试使用
  8. &lt;正向/反向&gt;最大匹配算法(Java)
  9. 用shell脚本爬取网页信息
  10. 【设计模式】Template Method模式
  11. phpMyAdmin安装与配置(涉及LAMP配置)
  12. jenkins2 -pipeline 常用groovy脚本
  13. Vue.js-09:第九章 - 组件基础再探(data、props)
  14. 使用Task
  15. ZKW线段树入门
  16. Atitti 创业团队vs打工的团队 attilax总结
  17. Java过滤器Filter的使用详解
  18. 远程获得乐趣的 Linux 命令
  19. python爬取酒店信息练习
  20. ubuntu下桌面假死处理方法(非重启)

热门文章

  1. maven打包记录1
  2. java多线程回顾3:线程安全
  3. java map遍历方式及效率
  4. javascript 使用 setInterval 实现倒计时
  5. nyoj 513-A+B Problem IV (java BigDecimal, stripTrailingZeros, toPlainString)
  6. PhpStudy BackDoor2019 深度分析
  7. vue中自定义html文件的模板
  8. PHP中Session ID的实现原理分析
  9. STDN: Scale-Transferrable Object Detection论文总结
  10. ES6扩展运算符...