【发布时间】：2016-04-02 14:14:02
【问题描述】：
我正在解决这个关于斐波那契数列的 codechef 问题。它说数字是 1000 位，那么为什么它在扫描数组并将其存储在 unsigned long long int 中时不会导致测试人员的解决方案中的整数溢出。我不明白解决方案是如何工作的。以下是问题和测试人员的解决方案。
The Head Chef has been playing with Fibonacci numbers for long . He has learnt several tricks related to Fibonacci numbers . Now he wants to test his chefs in the skills .
A fibonacci number is defined by the recurrence :
f(n) = f(n-1) + f(n-2) for n > 2
and f(1) = 0
and f(2) = 1 .
Given a number A , determine if it is a fibonacci number.
Input
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.
The only line of each test case contains a single integer A denoting the number to be checked .
Output
For each test case, output a single line containing "YES" if the given number is a fibonacci number , otherwise output a single line containing "NO" .
Constraints
1 ≤ T ≤ 1000
1 ≤ number of digits in A ≤ 1000
The sum of number of digits in A in all test cases <= 10000.
Example
Input:
3
3
4
5
Output:
YES
NO
YES
**Tester's solution:**
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <set>
#include <cstring>
using namespace std;
int const mx = 6666;
set <unsigned long long> f;
unsigned long long fib[mx + 10];
char s[mx + 1];
int main(){
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
fib[0] = 0;
fib[1] = 1;
f.insert(1);
f.insert(0);
int i;
for (i = 2; i <= mx; i++){
fib[i] = fib[i - 1] + fib[i - 2];
f.insert(fib[i]);
}
int tc;
cin>>tc;
while (tc--){
unsigned long long n = 0, ten = 10;
cin>>s;
int len = strlen(s);
for (i = 0; i < len; i++){
char q = s[i];
unsigned long long a = q - '0';
n = n * ten + a;
}
if (f.find(n) == f.end()) printf("NO\n");
else printf("YES\n");
}
return 0;
}
【问题讨论】：
标签： fibonacci integer-overflow