B. National Project

B. National Project

Your company was appointed to lay new asphalt on the highway of length n. You know that every day you can either repair one unit of the highway (lay new asphalt over one unit of the highway) or skip repairing.

Skipping the repair is necessary because of the climate. The climate in your region is periodical: there are g days when the weather is good and if you lay new asphalt these days it becomes high-quality pavement; after that, the weather during the next b days is bad, and if you lay new asphalt these days it becomes low-quality pavement; again g good days, b bad days and so on.

You can be sure that you start repairing at the start of a good season, in other words, days 1,2,…,g are good.

You don’t really care about the quality of the highway, you just want to make sure that at least half of the highway will have high-quality pavement. For example, if the n=5 then at least 3 units of the highway should have high quality; if n=4 then at least 2 units should have high quality.

What is the minimum number of days is needed to finish the repair of the whole highway?

Input
The first line contains a single integer T (1≤T≤104) — the number of test cases.

Next T lines contain test cases — one per line. Each line contains three integers n, g and b (1≤n,g,b≤109) — the length of the highway and the number of good and bad days respectively.

Output
Print T integers — one per test case. For each test case, print the minimum number of days required to repair the whole highway if at least half of it should have high quality.

Example
inputCopy

3
5 1 1
8 10 10
1000000 1 1000000

outputCopy

5
8
499999500000

Note
In the first test case, you can just lay new asphalt each day, since days 1,3,5 are good.

In the second test case, you can also lay new asphalt each day, since days 1-8 are good.

#include<iostream>
using namespace std;
int main() {
	int t;
	cin >> t;//样例个数
	long long n, g, b;//n为总天数，g为好的天数，b为差的天数；
	long long sum=0;
	long long d=0;
	while (t--) {
		sum = 0;
		d = 0;
		cin >> n >> g >> b;
		sum = (n + 1) / 2;//sum为需要修好路的天数
			long long a = sum / g;//需要好的差的天数多少个循环
			if (sum % g == 0) {
				d = g * a + b * (a - 1);//如果刚好做完，那么后面的坏的天数不计
			}
			else {
				d = a * (g + b) + sum %g;//如果没有做完，那么剩下的天数一定少于好的天数的天数，那么加上余数就好了；
			}
			if (d < n)cout << n << endl;//注意最后的天数一定要大于或等于最开始给出的总的天数，呜呜呜孩子在这里卡了好久，一定要最后判断啊！！！
			else cout << d << endl;
	}
}