按需取余

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CF 1374A. Required Remainder

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。

Input

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.

Output

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.

Example

输入

7
7 5 12345
5 0 4
10 5 15
17 8 54321
499999993 9 1000000000
10 5 187
2 0 999999999

输出

12339
0
15
54306
999999995
185
999999998

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~

#include<bits/stdc++.h>
using 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;
}

按需取余

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原文地址：https://www.cnblogs.com/RioTian/p/13395004.html