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AcWing 717. 简单斐波那契原题链接中等

作者：

第一WA者金银花 , 2019-08-07 10:43:17 , 所有人可见 , 阅读 14965

297

124

其实数组都不用开(炒鸡短)

#include<stdio.h>
long long n,a,b=1,c;
int main(){
    scanf("%d",&n);
    while(n--){
        printf("%d ",a);
        c=a+b;
        a=b,b=c;
    }
}

附赠高精度版一份(求点赞)

#include<stdio.h>
#include<memory.h>
#pragma GCC optimize(3)
#define M 100001
int n,a[M],b[M],c[M],la=1,lb=1;
void add(){
    memcpy(c,b,sizeof(b));
    la=lb;
    for(int i=1;i<=lb;i++){
        b[i]+=a[i];
        if(b[i]>9&&i<lb) b[i]-=10,b[i+1]++;
    }
    if(b[lb]>9) b[lb]-=10,b[++lb]++;
    memcpy(a,c,sizeof(c));
}
int main(){
    scanf("%d",&n),b[1]=1;
    while(n--){
        for(int i=la;i;i--) printf("%d",a[i]);
        printf(" ");
        add();
    }
}

18 评论

ssk 2022-10-30 16:50

超级短的很奇妙

MelodyEcho 2022-01-25 15:35

用 python 生成器也挺方便的 (￣▽￣)／

n1, n2, n = 0, 1, 0
def fibonacci():
    global n1, n2, n
    while True:
        if (n == 0 or n == 1): yield n
        else:
            sum = n1 + n2; n1 = n2; n2 = sum
            yield sum

g = fibonacci()
freq = int(input())

for i in g:
    print(i, end=' '); n += 1
    if (n == freq): break

acwing_2234 2022-10-27 15:11

妙啊这想法高精度

越关山 2021-08-15 00:49

#include<cstdio>
int main(){ 
    int n;
    scanf("%d",&n);
    n<=1?printf("0 "):printf("0 1 "); 
    for(int i = 1,j = 1,x = 1;x<n-1;x++,i+=j,j=i-j) printf("%d ",i);
    return 0;
}

一个做法