Revenge of Fibonacci hdu 5018

Problem Description
In mathematical terms, the sequence Fn of Fibonacci numbers is defined by the recurrence relation
Fn = Fn-1 + Fn-2
with seed values F1 = 1; F2 = 1 (sequence A000045 in OEIS).
—Wikipedia

Today, Fibonacci takes revenge on you. Now the first two elements of Fibonacci sequence has been redefined as A and B. You have to check if C is in the new Fibonacci sequence.

Input
The first line contains a single integer T, indicating the number of test cases.

Each test case only contains three integers A, B and C.

[Technical Specification]

1. 1 <= T <= 100
2. 1 <= A, B, C <= 1 000 000 000

Output
For each test case, output “Yes” if C is in the new Fibonacci sequence, otherwise “No”.

Sample Input

3
2 3 5
2 3 6
2 2 110

Sample Output

Yes
No
Yes
Hint
For the third test case, the new Fibonacci sequence is: 2, 2, 4, 6, 10, 16, 26, 42, 68, 110…

Source
BestCoder Round #10

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map 去映射一下即可；

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 200005
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
typedef long long  ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const int mod = 10000007;
#define Mod 20100403
#define sq(x) (x)*(x)
#define eps 1e-10


inline int rd() {
	int x = 0;
	char c = getchar();
	bool f = false;
	while (!isdigit(c)) {
		if (c == '-') f = true;
		c = getchar();
	}
	while (isdigit(c)) {
		x = (x << 1) + (x << 3) + (c ^ 48);
		c = getchar();
	}
	return f ? -x : x;
}

ll gcd(ll a, ll b) {
	return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }

map<ll, int>Map;
ll fib[200];

int main()
{
	//ios::sync_with_stdio(false);
	
	int T; cin >> T;
	while (T--) {
		ll A, B, C;
		ms(fib);
		Map.clear();
		rdllt(A); rdllt(B); rdllt(C);
		fib[1] = A; fib[2] = B;
		Map[A] = 1; Map[B] = 1;
		for (int i = 3; i <= 46; i++) {
			fib[i] = fib[i - 1] + fib[i - 2];
			Map[fib[i]] = 1;
		}
		if (Map[C])cout << "Yes" << endl;
		else cout << "No" << endl;
	}

}