标签：integer frame def 斐波那契数 nsis val cti sign ted
Input
Each test case consists of one integer n in a single line where 0≤n≤50. The input is terminated by -1.
Output
Print out the answer in a single line for each test case.
Sample Input
3 4 5 -1
Sample Output
2 3 5
Hint
you can use 64bit integer: __int64
// 递归
1 #include<stdio.h> 2 3 long fibbonacci(int n) 4 { 5 if(n==0) return 0; 6 else if(n==1) return 1; 7 else return fibbonacci(n-1) + fibbonacci(n-2); 8 } 9 10 int main() 11 { 12 int n; 13 while(scanf("%d", &n), n!=-1) 14 printf("%d\n", fibbonacci(n)); 15 return 0; 16 }
//
| int/long | __int64 |
|
signed: -2^31 ~ 2^31-1 ~ 2.1*10^9 unsigned: 0 ~ 2^32-1 ~ 4.29*10^9 |
signed:-2^63 ~ 2^63-1 ~ 9.2*10^18 unsigned: 0 ~ 2^64-1 ~ 1.8*10^19 |
// signed: scanf("%I64d",&a); printf("%I64d",a);
unsigned: scanf("%I64u",&a); printf("%I64u",a);
// 说明：
1、int64不能用作为循环变量
2、int64的操作速度较慢
1 #include<stdio.h> 2 3 __int64 fibbonacci(int n) 4 { 5 __int64 x1=0, x2=1, x3=0; 6 int i; 7 for(i=2;i<=n;i++) 8 { 9 x3=x1+x2; 10 x1=x2; x2=x3; 11 } 12 if(x3) return x3; 13 else 14 { 15 if(n) return x2; 16 else return x1; 17 } 18 } 19 20 int main() 21 { 22 int n; __int64 i; 23 while(scanf("%d", &n), n!=-1) 24 { 25 i=fibbonacci(n); 26 printf("%I64d\n", i); /* __intxx io格式 */ 27 } 28 return 0; 29 }
// 从第47个斐波那契数开始，其大小超过int范围
标签：integer frame def 斐波那契数 nsis val cti sign ted
原文地址：https://www.cnblogs.com/goldenretriever/p/10356654.html
