标签：sqrt put 找规律 void for his double matrix ble
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 = 1000000000;
constexpr int base_digits = 9;
struct bigint{
vector<int> z;
int sign;
bigint() : sign(1) {}
bigint(long long v) { *this = v; }
bigint& operator=(long long v)
{
sign = v < 0 ? -1 : 1;
v*=sign;
z.clear();
for(; v > 0; 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 = 0, carry = 0; i < other.z.size() || carry; ++i)
{
if(i==z.size()) z.push_back(0);
z[i] += carry + (i < other.z.size() ? other.z[i] : 0);
carry = z[i] >= base;
if(carry) z[i] -= base;
}
}
else if (other != 0 /* 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 == 1 && *this >= other || sign == -1 && *this <= other)
{
for (int i = 0, carry = 0; i < other.z.size() || carry; ++i)
{
z[i] -= carry + (i < other.z.size() ? other.z[i] : 0);
carry = z[i] < 0;
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<0) sign=-sign,v=-v;
for(int i=0,carry=0;i<z.size() || carry;++i)
{
if(i==z.size()) z.push_back(0);
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() + 1);
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() - 1; i >= 0; i--)
{
r*=base; r+=a.z[i];
int s1 = b.z.size() < r.z.size() ? r.z[b.z.size()] : 0;
int s2 = b.z.size() - 1 < r.z.size() ? r.z[b.z.size() - 1] : 0;
int d = (int)(((long long)s1 * base + s2) / b.z.back());
r -= b * d;
while(r < 0) 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()%2==1) a.z.push_back(0);
int n = a.z.size();
int firstDigit = (int)::sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
int norm = base / (firstDigit + 1);
a *= norm;
a *= norm;
while(a.z.empty()||a.z.size()%2==1) a.z.push_back(0);
bigint r = (long long)a.z[n - 1] * base + a.z[n - 2];
firstDigit = (int)::sqrt((double)a.z[n - 1] * base + a.z[n - 2]);
int q = firstDigit;
bigint res;
for (int j = n / 2 - 1; j >= 0; j--)
{
for(;;--q)
{
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);
if(r1>=0) { r=r1; break; }
}
res*=base;res+=q;
if(j>0)
{
int d1 = res.z.size() + 2 < r.z.size() ? r.z[res.z.size() + 2] : 0;
int d2 = res.z.size() + 1 < r.z.size() ? r.z[res.z.size() + 1] : 0;
int d3 = res.z.size() < r.z.size() ? r.z[res.z.size()]:0;
q = (int)(((long long)d1*base*base+(long long)d2*base+d3)/(firstDigit*2));
}
}
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<0) sign=-sign,v=-v;
for (int i = (int)z.size() - 1, rem = 0; i >= 0; --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<0) v=-v;
int m=0;
for(int i=(int)z.size()-1;i>=0;--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() - 1; i >= 0; 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() == 0) z.pop_back();
if(z.empty()) sign = 1;
}
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 == 1 ? *this : -*this;
}
long long longValue() const
{
long long res = 0;
for(int i = (int)z.size() - 1; i >= 0; 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 = 1;
z.clear();
int pos = 0;
while(pos < s.size() && (s[pos] == ‘-‘ || s[pos] == ‘+‘))
{
if(s[pos] == ‘-‘) sign = -sign;
++pos;
}
for(int i=(int)s.size()-1;i>=pos;i-=base_digits)
{
int x=0;
for(int j=max(pos,i-base_digits+1);j<=i;j++) x=x*10+s[j]-‘0‘;
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 == -1) stream << ‘-‘;
stream << (v.z.empty() ? 0 : v.z.back());
for(int i = (int)v.z.size() - 2; i >= 0; --i)
stream << setw(base_digits) << setfill(‘0‘) << 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) + 1);
p[0] = 1;
for(int i=1;i<p.size();i++) p[i]=p[i-1]*10;
vector<int> res;
long long cur = 0;
int cur_digits = 0;
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()==0)
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 <= 32)
{
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
res[i + j] += a[i] * b[j];
return res;
}
int k = n >> 1;
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=0;i<k;i++) a2[i]+=a1[i];
for(int i=0;i<k;i++) b2[i]+=b1[i];
vll r = karatsubaMultiply(a2, b2);
for(int i=0;i<a1b1.size();i++) r[i]-=a1b1[i];
for(int i=0;i<a2b2.size();i++) r[i]-=a2b2[i];
for(int i=0;i<r.size();i++) res[i+k]+=r[i];
for(int i=0;i<a1b1.size();i++) res[i]+=a1b1[i];
for(int i = 0;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,6);
vector<int> b6=convert_base(v.z,base_digits,6);
vll a(a6.begin(),a6.end());
vll b(b6.begin(),b6.end());
while(a.size()<b.size()) a.push_back(0);
while(b.size()<a.size()) b.push_back(0);
while(a.size()&(a.size()-1)) a.push_back(0),b.push_back(0);
vll c=karatsubaMultiply(a, b);
bigint res;
res.sign = sign * v.sign;
for (int i = 0, carry = 0; i < c.size(); i++)
{
long long cur = c[i] + carry;
res.z.push_back((int)(cur % 1000000));
carry = (int)(cur / 1000000);
}
res.z = convert_base(res.z, 6, base_digits);
res.trim();
return res;
}
};
bigint qpow(bigint a,bigint b){
bigint ans=1;
while(b!=0){
if(b%2){
ans= ans*a;
}
b/=2;
a= a*a;
}
return ans;
}
struct Matrix
{
bigint a[2][2];
Matrix()
{
rep(i,0,2){
rep(j,0,2){
a[i][j]=0;
}
}
}
Matrix operator * (const Matrix y)
{
Matrix ans;
for(int i = 0; i < 2; i++)
for(int j = 0; j < 2; j++)
for(int k = 0; k < 2; 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=0;i<2;i++)
for(int j=0;j<2;j++)
a[i][j]=y.a[i][j];
}
Matrix operator *= (const Matrix y)
{
Matrix ans;
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
for(int k=0;k<2;k++)
ans.a[i][j] = ans.a[i][j] + (a[i][k]*y.a[k][j]);
for(int i=0;i<2;i++)
for(int j=0;j<2;j++)
a[i][j]=ans.a[i][j];
}
};
Matrix qpow(bigint x)
{
Matrix ans;
ans.a[0][0]=ans.a[1][1]=1; //单位矩阵
Matrix mul;
mul.a[0][0]=4;
mul.a[0][1]=-1;
mul.a[1][0]=1;
mul.a[1][1]=0;
while(x!=0)
{
if(x%2!=0)
ans = ans*mul;
mul = mul* mul;
x/=2;
}
return ans;
}
bigint ans[1005];
void solve(){
ans[0]=(bigint)4;
ans[1]=(bigint)14;
ans[2]=(bigint)52;
rep(i,2,200){
ans[i]=(bigint)4*ans[i-1]-ans[i-2];
// cout<<ans[i]<<endl;
}
}
int main()
{
solve();
int t;cin>>t;
while(t--){
bigint n;
cin>>n;
rep(i,0,200){
if(ans[i]>=n){
cout<<ans[i]<<endl;
break;
}
}
}
}
2017 ACM/ICPC 沈阳 F题 Heron and his triangle
标签：sqrt put 找规律 void for his double matrix ble
原文地址：https://www.cnblogs.com/songorz/p/9784961.html
