Description：
In mathematics, the greatest common divisor ( g c d gcd 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 8 8 and 12 12 12 is 4 4 4.—Wikipedia
BrotherK and Ery like playing mathematic games. Today, they are playing a game with G C D GCD GCD.
BrotherK has an array
A
A
A with
N
N
N elements:
A
1
A
N
A_{1} ~ A_{N}
A1 AN, each element is a integer in
[
1
,
1
0
9
]
[1, 10^9]
[1,109]. Ery has Q questions, the i-th question is to calculate
G
C
D
(
A
L
i
,
A
L
i
+
1
,
A
L
i
+
2
,
.
.
.
,
A
R
i
)
GCD(A_{Li}, A_{Li+1}, A_{Li+2}, ..., A_{Ri})
GCD(ALi,ALi+1,ALi+2,...,ARi), and BrotherK will tell her the answer.
BrotherK feels tired after he has answered Q Q Q questions, so Ery can only play with herself, but she don’t know any elements in array A A A Fortunately, Ery remembered all her questions and BrotherK’s answer, now she wants to recovery the array A A A.
Input
The first line contains a single integer T T T, indicating the number of test cases.
Each test case begins with two integers N , Q N, Q N,Q, indicating the number of array A A A, and the number of Ery’s questions. Following Q Q Q lines, each line contains three integers L i , R i L_{i}, R_{i} Li,Ri and Ansi, describing the question and BrotherK’s answer.
T T T is about 10 10 10
2 ≤ N Q ≤ 1000 2 ≤ N Q ≤ 1000 2≤NQ≤1000
1 ≤ L i < R i ≤ N 1 ≤ L_{i} < R_{i} ≤ N 1≤Li<Ri≤N
1 ≤ A n s i ≤ 1 0 9 1 ≤ A_{nsi} ≤ 10^9 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 N N integer, i − t h i-th i−th integer is A i A_{i} 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
题意：
已知一个序列若干区间内数字的最大公约数，复原出原序列，输出最小满足条件的序列。
初始设序列上所有数字均为1，对每个区间上每个数字，变成原数字与公约数的最小公倍数，最后再检查一次结果序列每个区间是否满足。
AC代码：
#include <cstdio>
#include <vector>
#include <queue>
#include <cstring>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <iostream>
#include <algorithm>
#include <iomanip>
#include <stack>
#include <queue>
using namespace std;
#define sd(n) scanf("%d", &n)
#define sdd(n, m) scanf("%d%d", &n, &m)
#define sddd(n, m, k) scanf("%d%d%d", &n, &m, &k)
#define pd(n) printf("%d\n", n)
#define pc(n) printf("%c", n)
#define pdd(n, m) printf("%d %d", n, m)
#define pld(n) printf("%lld\n", n)
#define pldd(n, m) printf("%lld %lld\n", n, m)
#define sld(n) scanf("%lld", &n)
#define sldd(n, m) scanf("%lld%lld", &n, &m)
#define slddd(n, m, k) scanf("%lld%lld%lld", &n, &m, &k)
#define sf(n) scanf("%lf", &n)
#define sc(n) scanf("%c", &n)
#define sff(n, m) scanf("%lf%lf", &n, &m)
#define sfff(n, m, k) scanf("%lf%lf%lf", &n, &m, &k)
#define ss(str) scanf("%s", str)
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, a, n) for (int i = n; i >= a; i--)
#define mem(a, n) memset(a, n, sizeof(a))
#define debug(x) cout << #x << ": " << x << endl
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define fi first
#define se second
#define mod(x) ((x) % MOD)
#define gcd(a, b) __gcd(a, b)
#define lowbit(x) (x & -x)
typedef pair<int, int> PII;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
const int MOD = 1e9 + 7;
const double eps = 1e-9;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const int inf = 0x3f3f3f3f;
inline int read()
{
int ret = 0, sgn = 1;
char ch = getchar();
while (ch < '0' || ch > '9')
{
if (ch == '-')
sgn = -1;
ch = getchar();
}
while (ch >= '0' && ch <= '9')
{
ret = ret * 10 + ch - '0';
ch = getchar();
}
return ret * sgn;
}
inline void Out(int a) //Êä³öÍâ¹Ò
{
if (a > 9)
Out(a / 10);
putchar(a % 10 + '0');
}
ll gcd(ll a, ll b)
{
return b == 0 ? a : gcd(b, a % b);
}
ll lcm(ll a, ll b)
{
return a * b / gcd(a, b);
}
///快速幂m^k%mod
ll qpow(ll a, ll b, ll mod)
{
if (a >= mod)
a = a % mod + mod;
ll ans = 1;
while (b)
{
if (b & 1)
{
ans = ans * a;
if (ans >= mod)
ans = ans % mod + mod;
}
a *= a;
if (a >= mod)
a = a % mod + mod;
b >>= 1;
}
return ans;
}
// 快速幂求逆元
int Fermat(int a, int p) //费马求a关于b的逆元
{
return qpow(a, p - 2, p);
}
///扩展欧几里得
int exgcd(int a, int b, int &x, int &y)
{
if (b == 0)
{
x = 1;
y = 0;
return a;
}
int g = exgcd(b, a % b, x, y);
int t = x;
x = y;
y = t - a / b * y;
return g;
}
///使用ecgcd求a的逆元x
int mod_reverse(int a, int p)
{
int d, x, y;
d = exgcd(a, p, x, y);
if (d == 1)
return (x % p + p) % p;
else
return -1;
}
///中国剩余定理模板0
ll china(int a[], int b[], int n) //a[]为除数，b[]为余数
{
int M = 1, y, x = 0;
for (int i = 0; i < n; ++i) //算出它们累乘的结果
M *= a[i];
for (int i = 0; i < n; ++i)
{
int w = M / a[i];
int tx = 0;
int t = exgcd(w, a[i], tx, y); //计算逆元
x = (x + w * (b[i] / t) * x) % M;
}
return (x + M) % M;
}
ll ans[2122], l[1010], r[1010], a[1011];
int n;
int main()
{
int t;
sd(t);
while (t--)
{
int q, flag = 0;
sdd(n, q);
rep(i, 1, n)
ans[i] = 1;
rep(k, 1, q)
{
slddd(l[k], r[k], a[k]);
rep(i, l[k], r[k])
{
ans[i] = lcm(ans[i], a[k]);
if (ans[i] >= 1e9)
flag = 1;
}
}
rep(k, 1, q)
{
ll res = ans[l[k]];
rep(i, l[k] + 1, r[k])
res = gcd(res, ans[i]);
if (res != a[k])
flag = 1;
}
if (flag)
puts("Stupid BrotherK!");
else
{
rep(i, 1, n)
{
printf("%lld ", ans[i]);
}
cout << endl;
}
}
return 0;
}