Codeforces #657 Problem B

标签：while   cst   cep   copy   pre   stream   contains   差值   The   

题目

Pasha loves to send strictly positive integers to his friends. Pasha cares about security, therefore when he wants to send an integer n, he encrypts it in the following way: he picks three integers a, b and c such that l≤a,b,c≤r, and then he computes the encrypted value m=n?a+b?c.

Unfortunately, an adversary intercepted the values l, r and m. Is it possible to recover the original values of a, b and c from this information? More formally, you are asked to find any values of a, b and c such that

a, b and c are integers,
l≤a,b,c≤r,
there exists a strictly positive integer n, such that n?a+b?c=m.
Input
The first line contains the only integer t (1≤t≤20) — the number of test cases. The following t lines describe one test case each.

Each test case consists of three integers l, r and m (1≤l≤r≤500000, 1≤m≤1010). The numbers are such that the answer to the problem exists.

Output
For each test case output three integers a, b and c such that, l≤a,b,c≤r and there exists a strictly positive integer n such that n?a+b?c=m. It is guaranteed that there is at least one possible solution, and you can output any possible combination if there are multiple solutions.

Example
inputCopy
2
4 6 13
2 3 1
outputCopy
4 6 5
2 2 3
Note
In the first example n=3 is possible, then n?4+6?5=13=m. Other possible solutions include: a=4, b=5, c=4 (when n=3); a=5, b=4, c=6 (when n=3); a=6, b=6, c=5 (when n=2); a=6, b=5, c=4 (when n=2).

In the second example the only possible case is n=1: in this case n?2+2?3=1=m. Note that, n=0 is not possible, since in that case n is not a strictly positive integer.

思路

好像是个数学题。首先我们把式子化简，左式为n，然后枚举右边的分母a（l-r），其中如果a可以被m整除那么直接可以输出，其余两位相等即可。其他的话我们需要考虑两种情况，就是b-c是否大于0，如果大于0，那么我们需要找到差值是m%i，然后各加上l满足条件即可，另外一种情况可以视作m+（c-b），和另外一种情况一样的考虑方法，只不过这里c会比b大，那么我们给c加上一个i就可以了。讲到好绕还是看代码把。

代码实现

#include<cstdio>
#include<iostream>
#include<queue>
#include<vector>
#include<cstring>
#include<map>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long ll;
typedef pair <int ,int> PII;
int main () {
    ll t,l,r,m,a,b,c;
    cin>>t;
    while (t--) {
        cin>>l>>r>>m;
        for (int i=l;i<=r;i++) {
            if (m%i==0) {
                a=i;
                b=r;
                c=r;
                break;
            }
            else {
              if (m%i<=(r-l)&&m-(m%i)>0) {
                a=i;
                b=l+m%i;
                c=l;
                break;
            }
              else if (i-m%i<=(r-l)) {
                  a=i;
                  b=l;
                  c=l+i-m%i;
              }
        }
      }
      cout<<a<<" "<<b<<" "<<c<<endl;
    }
    return  0;
}

Codeforces #657 Problem B

标签：while   cst   cep   copy   pre   stream   contains   差值   The   

原文地址：https://www.cnblogs.com/hhlya/p/13341684.html