hdu 5223 GCD 思维题

GCD

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 506    Accepted Submission(s): 227

Problem Description

In mathematics, the greatest common divisor (gcd) of two or more integers, when at least one of them is not zero, is the largest positive integer that divides the numbers without a remainder. For example, the GCD of 8 and 12 is 4.---Wikipedia

BrotherK and Ery like playing mathematic games. Today, they are playing a game with GCD.

BrotherK has an array A with N elements: A1 ~ AN, each element is a integer in [1, 10^9]. Ery has Q questions, the i-th question is to calculate
GCD(ALi, ALi+1, ALi+2, ..., ARi), and BrotherK will tell her the answer.

BrotherK feels tired after he has answered Q questions, so Ery can only play with herself, but she don't know any elements in array A. Fortunately, Ery remembered all her questions and BrotherK's answer, now she wants to recovery the array A.

 

Input

The first line contains a single integer T, indicating the number of test cases.

Each test case begins with two integers N, Q, indicating the number of array A, and the number of Ery's questions. Following Q lines, each line contains three integers Li, Ri and Ansi, describing the question and BrotherK's answer.

T is about 10

2 ≤ N Q ≤ 1000

1 ≤ Li < Ri ≤ N

1 ≤ Ansi ≤ 109

 

Output

For each test, print one line.

If Ery can't find any array satisfy all her question and BrotherK's answer, print "Stupid BrotherK!" (without quotation marks). Otherwise, print N integer, i-th integer is Ai.

If there are many solutions, you should print the one with minimal sum of elements. If there are still many solutions, print any of them.

 

Sample Input

2
2 2
1 2 1
1 2 2
2 1
1 2 2

 

Sample Output

Stupid BrotherK!
2 2
#include <iostream>
#include<stdio.h>
#include<algorithm>
using namespace std;
struct node
{
    int l,r;
    int gcd;
}GCD[1000+10];
int a[1000+10];
int gcd(int a,int b)
{
    if(b==0)
        return a;
    return gcd(b,a%b);
}
int lcm(int a,int b)
{
    return a/gcd(a,b)*b;
}
int main()
{
     int t;
     cin>>t;
     while(t--)
     {
         int N,Q;
         cin>>N>>Q;
         for(int i=1;i<=N;i++)
            a[i]=1;
         for(int i=1;i<=Q;i++)
         {
             int l,r,gcd;
             scanf("%d%d%d",&l,&r,&gcd);
             GCD[i].l=l;GCD[i].r=r;GCD[i].gcd=gcd;
             for(int j=GCD[i].l;j<=GCD[i].r;j++)
                 a[j]=lcm(a[j],GCD[i].gcd);//先按照规则把每个数求出来
         }
         int ok=1;
         for(int i=1;i<=Q;i++)
         {
             int gcd2;
             gcd2=a[GCD[i].l];
              for(int j=GCD[i].l;j<=GCD[i].r;j++)
                gcd2=gcd(gcd2,a[j]);
             if(gcd2!=GCD[i].gcd)//验证每个查询区间
             {
                 ok=0;
                 break;
             }
         }
         if(ok==0)
         {
             printf("Stupid BrotherK!\n");
         }
         else
         {
             for(int i=1;i<N;i++)
                cout<<a[i]<<" ";
             cout<<a[N]<<endl;
         }

     }
}