HDU 5223 - GCD (思维)

题意

告诉每个区间的gcd，问能不能有这样一个数组。

思路

因为数据是1e3的级别，所以可以采取枚举的方式。

先按照题目给的情况，计算出每个点的答案。每个gcd相当于这个点有一个因子，那么只要求lcm就好了。

然后计算出所有区间的gcd。

然后判断是否和题目给出的gcd一样。

代码

#include <stack>#include <cstdio>#include <list>#include <cassert>#include <set>#include <fstream>#include <iostream>#include <string>#include <vector>#include <queue>#include <functional>#include <cstring>#include <algorithm>#include <cctype>#include <string>#include <map>#include <cmath>#include <ext/pb_ds/assoc_container.hpp>#include <ext/pb_ds/hash_policy.hpp>using namespace std;#define LL long long#define ULL unsigned long long#define SZ(x) (int)x.size()#define Lowbit(x) ((x) & (-x))#define MP(a, b) make_pair(a, b)#define MS(arr, num) memset(arr, num, sizeof(arr))#define PB push_back#define X first#define Y second#define ROP freopen("input.txt", "r", stdin);#define MID(a, b) (a + ((b - a) >> 1))#define LC rt << 1, l, mid#define RC rt << 1|1, mid + 1, r#define LRT rt << 1#define RRT rt << 1|1#define FOR(i, a, b) for (int i=(a); (i) < (b); (i)++)const double PI = acos(-1.0);const int INF = 0x3f3f3f3f;const double eps = 1e-8;const int MAXN = 1000+10;const int MOD = 1e9;const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };const int seed = 131;int cases = 0;typedef pair<int, int> pii;struct CMD{    int l, r, gcd;}cmd[MAXN];int arr[MAXN], range_gcd[MAXN][MAXN];bool flag;int lcm(LL a, LL b){    LL g = __gcd(a, b);    LL ret = a / g * b;    if (ret > (int)(1e9))    //lcm不能大于1e9    {        flag = true;        return 0;    }    return ret;}int main(){    //ROP;    int T;    scanf("%d", &T);    while (T--)    {        flag = false;        int n, k;        scanf("%d%d", &n, &k);        fill(arr+1, arr+1+n, 1);        FOR(i, 0, k)        {            int a, b, gcd;            scanf("%d%d%d", &cmd[i].l, &cmd[i].r, &cmd[i].gcd);        }        for (int i = 1; i <= n; i++)        {            if (flag) break;            FOR(j, 0, k)            {                if (i >= cmd[j].l && i <= cmd[j].r) arr[i] = lcm(arr[i], cmd[j].gcd);                if (flag) break;            }        }        if (flag) { puts("Stupid BrotherK!"); continue; }        //计算各个区间的gcd        for (int i = 1; i <= n; i++)        {            range_gcd[i][i] = arr[i];            for (int j = i+1; j <= n; j++) range_gcd[i][j] = __gcd(range_gcd[i][j-1], arr[j]);        }        flag = false;        FOR(i, 0, k)        {            int l = cmd[i].l, r = cmd[i].r;            if (range_gcd[l][r] != cmd[i].gcd)            {                flag = true;                break;            }            if (flag) break;        }        if (flag) puts("Stupid BrotherK!");        else        {            FOR(i, 1, n) printf("%d ", arr[i]);            printf("%d\n", arr[n]);        }    }    return 0;}