m=n⋅a+b−c(n为任意正整数),给出m的值a,b,c的范围l,r(l<=a,b,c<=r),求出a,b,c。
由推倒知0<=|b-c|<=r-l,然后枚举看能否有m%a或者a-m%a能够>=0或<=r-l即可。
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
const int N=2e5+7;
const double eps=1e-8;
const int mod=1e9+7;
const int inf=0x7fffffff;
const double pi=3.1415926535;
signed main()
{
IOS;
int t;
cin>>t;
while(t--)
{
int a,b,m;
cin>>a>>b>>m;
for(int i=a;i<=b;i++)
{
int ret=m%i,op=m/i;
int res=i-ret;
if(ret<=b-a&&op)
{
cout<<i<<" "<<a+ret<<" "<<a<<endl;
break;
}
else if(res<=b-a)
{
cout<<i<<" "<<b-res<<" "<<b<<endl;break;
}
}
}
return 0;
}
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