time limit per test
2 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output
Leo has developed a new programming language C+=. In C+=, integer variables can only be changed with a "+=" operation that adds the right-hand side value to the left-hand side variable. For example, performing "a += b" when a = 22, b = 33 changes the value of a to 55 (the value of b does not change).
In a prototype program Leo has two integer variables a and b, initialized with some positive values. He can perform any number of operations "a += b" or "b += a". Leo wants to test handling large integers, so he wants to make the value of either a or b strictly greater than a given value nn. What is the smallest number of operations he has to perform?
Input
The first line contains a single integer TT (1≤T≤1001≤T≤100) — the number of test cases.
Each of the following TT lines describes a single test case, and contains three integers a,b,na,b,n (1≤a,b≤n≤1091≤a,b≤n≤109) — initial values of a and b, and the value one of the variables has to exceed, respectively.
Output
For each test case print a single integer — the smallest number of operations needed. Separate answers with line breaks.
Example
input
Copy
2 1 2 3 5 4 100
output
Copy
2 7
Note
In the first case we cannot make a variable exceed 33 in one operation. One way of achieving this in two operations is to perform "b += a" twice.
解題說明：此題採用貪心算法，保證數字之和經過迭代後超過n即可，每次加選擇a和b中最大的在前面。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int t;
long int a, b, n, count;
scanf("%d", &t);
while (t--)
{
scanf("%ld%ld%ld", &a, &b, &n);
count = 0;
while (!(a>n || b>n))
{
if (a < b)
{
a += b;
}
else
{
b += a;
}
count++;
}
printf("%ld\n", count);
}
return 0;
}