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#include <algorithm> // max
#include <cassert>   // assert
#include <cstdio>    // printf,sprintf
#include <cstring>   // strlen
#include <iostream>  // cin,cout
#include <string>    // string類
#include <vector>    // vector類
using namespace std;

struct BigInteger {
    typedef unsigned long long LL;

    static const int BASE = 100000000;
    static const int WIDTH = 8;
    vector<int> s;

    BigInteger& clean(){while(!s.back()&&s.size()>1)s.pop_back(); return *this;}
    BigInteger(LL num = 0) {*this = num;}
    BigInteger(string s) {*this = s;}
 	BigInteger& operator = (long long num) {      							
        s.clear();
        do {
            s.push_back(num % BASE);
            num /= BASE;
        } while (num > 0);
        return *this;
    }
    BigInteger& operator = (const string& str) {
        s.clear();
        int x, len = (str.length() - 1) / WIDTH + 1;
        for (int i = 0; i < len; i++) {
            int end = str.length() - i*WIDTH;
            int start = max(0, end - WIDTH);
            sscanf(str.substr(start,end-start).c_str(), "%d", &x);
            s.push_back(x);
        }
        return (*this).clean();
    }

    BigInteger operator + (const BigInteger& b) const {
        BigInteger c; c.s.clear();
        for (int i = 0, g = 0; ; i++) {
            if (g == 0 && i >= s.size() && i >= b.s.size()) break;
            int x = g;
            if (i < s.size()) x += s[i];
            if (i < b.s.size()) x += b.s[i];
            c.s.push_back(x % BASE);
            g = x / BASE;
        }
        return c;
    }
    BigInteger operator - (const BigInteger& b) const {
        assert(b <= *this); // 減數不能大於被減數
        BigInteger c; c.s.clear();
        for (int i = 0, g = 0; ; i++) {
            if (g == 0 && i >= s.size() && i >= b.s.size()) break;
            int x = s[i] + g;
            if (i < b.s.size()) x -= b.s[i];
            if (x < 0) {g = -1; x += BASE;} else g = 0;
            c.s.push_back(x);
        }
        return c.clean();
    }
    BigInteger operator * (const BigInteger& b) const {
        int i, j; LL g;
        vector<LL> v(s.size()+b.s.size(), 0);
        BigInteger c; c.s.clear();
        for(i=0;i<s.size();i++) for(j=0;j<b.s.size();j++) v[i+j]+=LL(s[i])*b.s[j];
        for (i = 0, g = 0; ; i++) {
            if (g ==0 && i >= v.size()) break;
            LL x = v[i] + g;
            c.s.push_back(x % BASE);
            g = x / BASE;
        }
        return c.clean();
    }
    BigInteger operator / (const BigInteger& b) const {
        assert(b > 0);  // 除數必須大於0
        BigInteger c = *this;       // 商:主要是讓c.s和(*this).s的vector一樣大
        BigInteger m;               // 餘數:初始化為0
        for (int i = s.size()-1; i >= 0; i--) {
            m = m*BASE + s[i];
            c.s[i] = bsearch(b, m);
			m -= b*c.s[i];
        }
        return c.clean();
    }
    BigInteger operator % (const BigInteger& b) const { //方法與除法相同
        BigInteger c = *this;
        BigInteger m;
        for (int i = s.size()-1; i >= 0; i--) {
            m = m*BASE + s[i];
            c.s[i] = bsearch(b, m);
            m -= b*c.s[i];
        }
        return m;
    }
    // 二分法找出滿足bx<=m的最大的x
    int bsearch(const BigInteger& b, const BigInteger& m) const{
        int L = 0, R = BASE-1, x;
        while (1) {
            x = (L+R)>>1;
            if (b*x<=m) {if (b*(x+1)>m) return x; else L = x;}
            else R = x;
        }
    }
    BigInteger& operator += (const BigInteger& b) {*this = *this + b; return *this;}
    BigInteger& operator -= (const BigInteger& b) {*this = *this - b; return *this;}
    BigInteger& operator *= (const BigInteger& b) {*this = *this * b; return *this;}
    BigInteger& operator /= (const BigInteger& b) {*this = *this / b; return *this;}
    BigInteger& operator %= (const BigInteger& b) {*this = *this % b; return *this;}

    bool operator < (const BigInteger& b) const {
        if (s.size() != b.s.size()) return s.size() < b.s.size();
        for (int i = s.size()-1; i >= 0; i--)
            if (s[i] != b.s[i]) return s[i] < b.s[i];
        return false;
    }
    bool operator >(const BigInteger& b) const{return b < *this;}
    bool operator<=(const BigInteger& b) const{return !(b < *this);}
    bool operator>=(const BigInteger& b) const{return !(*this < b);}
    bool operator!=(const BigInteger& b) const{return b < *this || *this < b;}
    bool operator==(const BigInteger& b) const{return !(b < *this) && !(b > *this);}
};

 ostream& operator << (ostream& out, const BigInteger& x) {
    out << x.s.back();
    for (int i = x.s.size()-2; i >= 0; i--) {
        char buf[20];
        sprintf(buf, "%08d", x.s[i]);
        for (int j = 0; j < strlen(buf); j++) out << buf[j];
    }
    return out;
}

  istream& operator >> (istream& in, BigInteger& x) {
    string s;
    if (!(in >> s)) return in;
    x = s;
    return in;
}
//BigInteger num1[101],num2[101],num3[101];
int main(){
	BigInteger a("15762679542071167858843489"),b("24332675219681431451788241"),c( "28992087708416717612934417");
	BigInteger i,j,k;
	while(cin>>i>>j>>k){
		cout<<a*i+b*j+c*k<<endl;
	}

//註釋的這些語句是為了找到A99是由多少個 a0 、 a1 、a2 組成。 
 
//	num1[1]=1;
//	num2[2]=1;
//	num3[3]=1;
//	for(int i=4;i<=100;i++){
//		num1[i]=num1[i-1]+num1[i-2]+num1[i-3];
//		num2[i]=num2[i-1]+num2[i-2]+num2[i-3];
//		num3[i]=num3[i-1]+num3[i-2]+num3[i-3];
//		cout<<num1[i]<<" "<<num2[i]<<" "<<num3[i]<<endl;
//	}
//	
	
	return 0;
}