# 11017: 【原1017】二哥养兔子

### 题目描述

author: 吴杰 原OJ链接：https://acm.sjtu.edu.cn/OnlineJudge-old/problem/1017

## Sample Input

0 1 1 11


## Sample Output

144


## FineArtz's solution

/* 二哥养兔子 */
#include <iostream>
#include <string>
using namespace std;

class BigInt{
friend BigInt operator +(const BigInt&, const BigInt&);
friend BigInt operator *(const BigInt&, const int&);
friend ostream& operator <<(ostream &, const BigInt&);
public:
//constructor
BigInt() = default;
BigInt(string);
BigInt(const BigInt&);

int getl() const { return len; }
private:
int data[20000] = {0};
int len = 1;
};

BigInt::BigInt(string s){
len = s.size();
for (int i = 1; i <= len; ++i)
data[i] = s[len - i] - '0';
}
BigInt::BigInt(const BigInt &b){
len = b.len;
for (int i = 1; i <= len; ++i)
data[i] = b.data[i];
}

BigInt operator +(const BigInt &a, const BigInt &b){
BigInt ans;
int l = (a.getl() > b.getl() ? a.getl() : b.getl());
for (int i = 1; i <= l; ++i)
ans.data[i] = a.data[i] + b.data[i];
int i = 1;
while (i <= l + 1){
if (ans.data[i] >= 10){
ans.data[i++] -= 10;
++ans.data[i];
}
else ++i;
}
ans.len = i;
while (ans.len > 1 && ans.data[ans.len] == 0) --ans.len;
return ans;
}
BigInt operator *(const BigInt &a, const int &b){
BigInt ans;
for (int i = 1; i <= a.len; ++i)
ans.data[i] = b * a.data[i];
ans.len = a.len;
for (int i = 1; i <= ans.len; ++i){
if (ans.data[i] / 10 != 0){
ans.data[i + 1] += ans.data[i] / 10;
ans.data[i] %= 10;
}
}
while (ans.data[ans.len + 1] > 0){
++ans.len;
if (ans.data[ans.len] >= 10){
ans.data[ans.len + 1] += ans.data[ans.len] / 10;
ans.data[ans.len] %= 10;
}
}
return ans;
}
ostream& operator <<(ostream &os, const BigInt &a){
for (int i = a.len; i >= 1; --i)
os << a.data[i];
return os;
}

BigInt one("1"), two, aft, ans("1");
int main(){
int a, b, c, n;
cin >> a >> b >> c >> n;
while (n--){
BigInt tmp(one * a + two * b + aft * c);
ans = ans + tmp;
aft = aft + two;
two = one;
one = tmp;
}
cout << ans << endl;
// cout << ans.getl() << endl;
}


## ligongzzz's solution

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<string>
#include<vector>

using namespace std;
struct bigint { // only positive number;
static const int BASE = 100000000;
static const int WIDTH = 8;
vector<int> s;
//value
bigint(long long num = 0) { *this = num; }
bigint operator = (long long num) {
s.clear();
do {
s.push_back(num%BASE);
num /= BASE;
} while (num > 0);
return *this;
}
bigint 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_s(str.substr(start, end - start).c_str(), "%d", &x);
s.push_back(x);
}
return *this;
}
//input&output
friend ostream& operator << (ostream &out, const bigint& x) {
out << x.s.back();
for (int i = x.s.size() - 2; i >= 0; i--) {
char buf[20];
sprintf_s(buf, "%08d", x.s[i]);
for (int j = 0; j < strlen(buf); j++) out << buf[j];
}
return out;
}
friend istream& operator >>(istream &in, bigint& x) {
string s;
if (!(in >> s)) return in;
x = s;
return in;
}
//compare
bool operator < (const bigint& 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;//equal
}
bool operator > (const bigint& b) const { return b < *this; }
bool operator <= (const bigint& b) const { return !(b < *this); }
bool operator >= (const bigint& b) const { return !(*this < b); }
bool operator != (const bigint& b) const { return b < *this || *this < b; }
bool operator == (const bigint& b) const { return !(b < *this) && !(*this < b); }
//calculate
bigint operator +(const bigint& b) const {
bigint 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;
}
bigint operator +=(const bigint& b) {
*this = *this + b;
return *this;
}
bigint operator -(const bigint& b) const {
bigint 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];
x += BASE;
c.s.push_back(x%BASE);
g = x / BASE - 1;
}
return c;
}
bigint operator * (const bigint& b) const {
bigint c;
c.s.clear();
bigint g = 0;
for (int i = 0;; i++) {
if (g.s.size() == 0 && i >= s.size() + b.s.size() - 1) break;
bigint x;
x.s.clear();
for (int j = 0; j < g.s.size(); j++) x.s.push_back(g.s[j]);
if (i < s.size() + b.s.size() - 1) {
for (int j = max(0, i - (int)s.size() + 1); j <= min(i, (int)b.s.size() - 1); j++) {
bigint t = (long long)b.s[j] * s[i - j];
x += t;
}
}
c.s.push_back(x.s[0]);
g.s.clear();
if (x.s.size() > 1) for (int j = 1; j < x.s.size(); j++) g.s.push_back(x.s[j]);
}
return c;
}
};

int main() {
bigint a, b, c;
int n;
bigint num;
bigint num1 = 1, num2 = 0, num3 = 0;

//输入
cin >> a >> b >> c >> n;

//计算
for (int i = 0; i < n; i++) {
bigint temp = num1 * a + num2 * b + num3 * c;
num3 += num2;
num2 = num1;
num1 = temp;
}

cout << num1 + num2 + num3 << endl;

return 0;
}


## yyong119's solution

#include <iostream>
int num[1005][2501];
int a,b,c,n,i,j;
int main(){
using namespace std;
cin>>a>>b>>c>>n;
num[4][0]=1; num[4][1]=1;
for (i=5; i<=n+4; i++){
for (j=1; j<=num[i-1][0]; j++) num[i][j]+=num[i-1][j];
for (j=1; j<=num[i-3][0]; j++) num[i][j]+=c*num[i-3][j];
for (j=1; j<=num[i-2][0]; j++) num[i][j]+=b*(num[i-2][j]-num[i-3][j]);
for (j=1; j<=num[i-1][0]; j++) num[i][j]+=a*(num[i-1][j]-num[i-2][j]);
for (j=1; j<=2500; j++){
num[i][j+1]+=num[i][j]/10;
num[i][j]=num[i][j]%10;
}
for (j=2500; j>=1; j--) if (num[i][j]){num[i][0]=j; break;}
}
for (i=num[n+4][0]; i>=1; i--) cout<<num[n+4][i];
return 0;
}