爱站程序员基地Required Remainder：
题目大意：（文末有原题）
给出一个整数n，输出0~n中 对x的余为y的最大整数；
思路：
要获得对x取余为y的最大整数，其实就是最多能包含多少个x（设最多为k个），并且n >= k * x + y，所以只需求出n/x再进行判断是否满足n >= k * x + y即可；
代码：
[code]#include <iostream>using namespace std;typedef long long ll;int main() {int t;cin >> t;while(t--) {ll x, y, n, s, z;cin >> x >> y >> n;s = (n / x) * x;if(n - s < y) {s -= x;}z = s + y;cout << z << endl;}return 0;}
原题：
题目：
You are given three integers x,y and n . Your task is to find the maximum integer k such that 0≤k≤n that k mod x=y , where mod is modulo operation. Many programming languages use percent operator % to implement it.
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.
输入：
The first line of the input contains one integer t (1≤t≤5⋅10^4) — 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≤10^9; 0≤y<x; y≤n≤10^9 ).
It can be shown that such kk always exists under the given constraints.
输出：
For each test case, print the answer — maximum non-negative integer k such that 0≤k≤n and k mod x = y . It is guaranteed that the answer always exists.
样例：
Input： Output:
7
7 5 12345 —————————- 12339
5 0 4 ———————————– 0
10 5 15 ——————————– 15
17 8 54321 ————————— 54306
499999993 9 1000000000 ——- 999999995
10 5 187 —————————— 185
2 0 999999999 ———————– 999999998Note
In the first test case of the example, the answer is 12339=7⋅1762+5 (thus, 12339mod7=5). It is obvious that there is no greater integer not exceeding 12345 which has the remainder 5 modulo 7.
