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You are given three integers x,y and n. Your task is to find the maximum integer k such that 0≤k≤n that kmodx=y, where mod is modulo operation. Many programming languages use percent operator % to implement it.
给定3个整数x,y和n，找到一个数k，使得
，并且
。
In other words, with given x,y and n you need to find the maximum possible integer from 0 to n that has the remainder y modulo x.
You have to answer t independent test cases. It is guaranteed that such k exists for each test case.
给定t组测试数据，每组数据保证存在这样的k。
The first line of the input contains one integer t (1≤t≤5⋅104) — the number of test cases. The next t lines contain test cases.
The only line of the test case contains three integers x,y and n (2≤x≤109; 0≤y<x; y≤n≤109).
It can be shown that such k always exists under the given constraints.
For each test case, print the answer — maximum non-negative integer k such that 0≤k≤n and kmodx=y. It is guaranteed that the answer always exists.
77 5 123455 0 410 5 1517 8 54321499999993 9 100000000010 5 1872 0 999999999
1233901554306999999995185999999998
比如输入x=7，y=5，n=12345，如何求得k，使得k<12345并且k%7=5呢？
朴素的做法：遍历n到1，看k%7是否等于5
换一种思路：考虑p%7是否等于0，如果等于0，p再加上5不就是答案k了吗。显然直接用n/b取底再乘以b即是最大的p。结果可以用公式表示如下：
当我们在解决一个问题的时候，首先看能不能将这个问题转换为相近的问题，间接的去思考最终结果，虽然是一个很小的题目，但是可以多花一点时间思考为什么。
每天学习一点点，你学会了吗，这是CF最简单的题目啦，加油~
#includeusing namespace std;typedef long long ll;int main() { int t; cin >> t; ll x, y, n; while (t--) { cin >> x >> y >> n; cout << (n - y) / x * x + y << endl; } return 0;}
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