UVA 10579 Fibonacci Numbers(高精度加法)

discription

一个斐波那契数列

f(1)=1,f(2)=1,f(n>2)=f(n−1)+f(n−2)

input

输入一个n

output

输出相应的斐波那契数列数字

Sample Input

3
100

Sample Output

2
354224848179261915075

注意：没有生成的斐波那契数列数字超过100位数字。

solution

高精度加法

code

#include <iostream>#include <cstdio>#include <algorithm>#include <vector>#include <cstring>#include <string>using namespace std;string add(string str1, string str2) //高精度加法{    string str;    int len1 = str1.length();    int len2 = str2.length();    //前面补0，弄成长度相同    if (len1 < len2)    {        for (int i = 1; i <= len2 - len1; i++)            str1 = "0" + str1;    }    else    {        for (int i = 1; i <= len1 - len2; i++)            str2 = "0" + str2;    }    len1 = str1.length();    int cf = 0;    int temp;    for (int i = len1 - 1; i >= 0; i--)    {        temp = str1[i] - '0' + str2[i] - '0' + cf;        cf = temp / 10;        temp %= 10;        str = char(temp + '0') + str;    }    if (cf != 0)        str = char(cf + '0') + str;    return str;}int main(){    // freopen("in.txt", "r", stdin);    int n;    while (~scanf("%d", &n))    {        string n1 = "1", n2 = "1", n3 = "0";        int cnt = n - 2;        for (; cnt > 0; cnt--)        {            n3 = add(n1, n2);            n1 = n2, n2 = n3;        }        if (n == 1 || n == 2)            printf("1\n");        else            cout << n3 << '\n';    }    return 0;}