(赛码"BestCoder"杯中国大学生程序设计冠军赛)GCD

GCD

 

 Accepts: 433

 

 Submissions: 1753

 Time Limit: 2000/1000 MS (Java/Others)

 

 Memory Limit: 131072/131072 K (Java/Others)

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<cstring>
using namespace std;
long long  v[1005],a[1005][3];

long long mm(long long  m,long long  n)
{
  long long  a,b,c;

        if( n > m) {
            a = n ;
            b = m ;
        }
        else {
            a = m ; b = n;
        }
        while ( b != 0 ) {  /* 最小公倍数 = m*n/GCD(m,n) */
            c = a % b ;
            a = b ;
            b = c ;
        }
    return a ;

}
long long  ma(long long  m,long long n){
    if(m==0)return n;
    else
return (m*n)/mm(m,n);
}
long long  gcd(long long  x,long long y){
long long  l=v[x];
for(int i=x+1;i<=y;i++)
l=mm(l,v[i]);

return l;
}
int main(){
int t;
cin>>t;
int n,m;
long long x,y,z;
while(t--)
{bool p=1;
memset(v,0,sizeof(v));
memset(a,0,sizeof(a));
 cin>>n>>m;
 for(int i=0;i<m;i++)
 {
     cin>>x>>y>>z;
     a[i][0]=x;a[i][1]=y;
     a[i][2]=z;
     for(int k=x;k<=y;k++){
     v[k]=ma(v[k],z);
     }
 }
int i;
for(int i=0;i<m;i++)
    if(gcd(a[i][0],a[i][1])!=a[i][2]){
        p=0;break;
   }

if(p==0)cout<<"Stupid BrotherK!";
else
{
if(v[1]==0)cout<<1;
else cout<<v[1];
for(int i=2;i<=n;i++)

    if(v[i]==0)cout<<" "<<1;
else cout<<" "<<v[i];
}

cout<<endl;
}
return 0;}

转载于:https://www.cnblogs.com/lengxia/p/4472249.html