uva12459

uva12459

分析了题目,就是求斐波那契数列的第N项,而且也告知了最大范围,即第80项,代码也非常简单

python版本代码

Fibo = []
for i in range(81):
    if i == 0 or i == 1:
        Fibo.append(1)
    else:
        Fibo.append(Fibo[i-2] + Fibo[i-1])

while True:
    a = int(input())
    if a == 0:
        break
    else:
        print(Fibo[a])

C/C++版本代码

#include <iostream>
#include<cstdio>
using namespace std;

//#define ZANGFONG

long long F[81];


int main()
{
    #ifdef ZANGFONG
    freopen("in.txt","r",stdin);
    freopen("out.txt","w",stdout);
    #endif // ZANGFONG
    int i,n;
    F[0] = F[1] = 1;
    for(i = 2; i < 81; i++)
    {
        F[i] = F[i-1] + F[i-2];
    }
    while(scanf("%d\n",&n) && n)
    {
        printf("%lld\n",F[n]);
    }

    return 0;
}

猜你喜欢

转载自blog.csdn.net/zangfong/article/details/82504002