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）
#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;
            }
        }

    }
}

　　

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

时间： 2024-10-11 21:14:57

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