斐波那契数列（大数加法）

题意：

求斐波那契的前10000项目

分析：

模拟竖式加法， 用string作为数字的储存形式

#include<bits/stdc++.h>
using namespace std;
string add1(string s1, string s2)
{
    if (s1 == "" && s2 == "")   return "0";
    if (s1 == "")   return s2;
    if (s2 == "")   return s1;
    string maxx = s1, minn = s2;
    if (s1.length() < s2.length()){
        maxx = s2;
        minn = s1;
    }
    int a = maxx.length() - 1, b = minn.length() - 1;
    for (int i = b; i >= 0; --i){
        maxx[a--] += minn[i] - '0'; //  a一直在减 ， 额外还要减个'0'
    }
    for (int i = maxx.length()-1; i > 0;--i){
        if (maxx[i] > '9'){
            maxx[i] -= 10;//注意这个是减10
            maxx[i - 1]++;
        }
    }
    if (maxx[0] > '9'){
        maxx[0] -= 10;
        maxx = '1' + maxx;
    }
    return maxx;
}
int main(){
    string a[10007];
    a[3] = "4" , a[4] = "7";
    for(int i = 5; i <= 10000; i++){
        a[i] = add1(a[i-1],a[i-2]);
    }
    int n;
    while(cin >> n){
        cout << a[n] << "\n";
    }
}