[codeforces 317]A. Perfect Pair

[codeforces 317]A. Perfect Pair

试题描述

Let us call a pair of integer numbers m-perfect, if at least one number in the pair is greater than or equal to m. Thus, the pairs (3, 3) and (0, 2) are 2-perfect while the pair (-1, 1) is not.

Two integers x, y are written on the blackboard. It is allowed to erase one of them and replace it with the sum of the numbers, (x + y).

What is the minimum number of such operations one has to perform in order to make the given pair of integers m-perfect?

输入

Single line of the input contains three integers x, y and m ( - 10¹⁸ ≤ x, y, m ≤ 10¹⁸).

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preffered to use the cin, cout streams or the %I64dspecifier.

输出

Print the minimum number of operations or "-1" (without quotes), if it is impossible to transform the given pair to the m-perfect one.

输入示例

-  

输出示例

数据规模及约定

见“输入”

题解

如果都是整数，那么不难发现增长率和斐波那契数列相同，就是指数级的，所以直接模拟即可。注意特判有负数的情况。如果只有一个负数，先把这个负数加成正的再模拟。如果两个都是负数，那么不可能再变大了。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
using namespace std;
#define LL long long

const int BufferSize = 1 << 16;
char buffer[BufferSize], *Head, *Tail;
inline char Getchar() {
    if(Head == Tail) {
        int l = fread(buffer, 1, BufferSize, stdin);
        Tail = (Head = buffer) + l;
    }
    return *Head++;
}
LL read() {
    LL x = 0, f = 1; char c = getchar();
    while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); }
    while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); }
    return x * f;
}

LL x, y, m;

int main() {
	x = read(); y = read(); m = read();

	if(x > y) swap(x, y);
	if(x <= 0 && y <= 0 && y < m) return puts("-1"), 0;
	if(y >= m) return puts("0"), 0;
	LL cnt = abs(x) % y ? abs(x) / y + 1 : abs(x) / y;
	x += cnt * y; if(x > y) swap(x, y);
	while(y < m) {
		x = x + y;
		if(x > y) swap(x, y);
		cnt++;
	}

	printf("%I64d\n", cnt);

	return 0;
}