题目

Fibonacci数列，定义如下：
f(1)=f(2)=1
f(n)=f(n-1)+f(n-2) n>=3。
计算第n项Fibonacci数值。

输入

输入第一行为一个整数N，接下来N行为整数Pi（1<=Pi<=1000）。

输出

输出为N行，每行为对应的f(Pi)。

样例输入

5
1
2
3
4
5
样例输出

1
1
2
3
5

分析与解答：

biginteger分析参见：
https://blog.csdn.net/mrcrack/article/details/53263235
代码：

#include<iostream>
#include<vector>
#include<cstring>
using namespace std;
struct BigInteger {static const int BASE = 100000000;static const int WIDTH = 8;vector<int> s;BigInteger(long long num = 0) { *this = num; } // 构造函数BigInteger operator = (long long num) { // 赋值运算符s.clear();do {s.push_back(num % BASE);num /= BASE;} while (num > 0);return *this;}BigInteger operator = (const string& str) { // 赋值运算符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;}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;}
};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;
}int main(){BigInteger a,b,ans[1005];ans[1]=1;ans[2]=1;for(int i=3;i<=1001;++i){ans[i]=ans[i-1]+ans[i-2];}int n;cin>>n;while(n--){int i;cin>>i;cout<<ans[i]<<endl;}
}