HDU 5214 Movie

Problem Description

Cloud and Miceren like watching movies. 

Today, they want to choose some wonderful scenes from a movie. A movie has 

N scenes can be chosen, and each scene is associate with an interval [

L, 

R]. 

L is the beginning time of the scene and 

R is the ending time. However, they can't choose two scenes which have overlapping intervals. (For example, scene with [1, 2] and scene with [2, 3], scene with [2, 5] and scene with[3, 4]). 

Now, can you tell them if they can choose such three scenes that any pair of them do not overlap? 

Since there are so many scenes that you can't get them in time, we will give you seven parameters 

N, L1, R1, a, b, c, d, and you can generate 

L1 ~ 

LN, 

R1 ~ 

RN

Input

T, indicating the number of test cases.

Each test case contains seven integers 

N, L1, R1, a, b, c, d, meaning that there are 

N scenes. The i-th scene's interval is [

Li, Ri]. 

L1 and 

R1 have been stated in input, and 

Li = (Li−1 ∗ a + b) mod 4294967296, Ri = (Ri−1 ∗ c + d) mod 4294967296. 

After all the intervals are generated, swap the i-th interval's 

Li and 

Ri if 

Li > Ri.

T is about 100.

1 ≤ N ≤ 10000000.

1 ≤ L1,R1 ≤ 2000000000.

1 ≤ a,b,c,d ≤ 1000000000.

The ratio of test cases with 

N > 100

Output

For each test, print one line.

If they can choose such three scenes, output "YES", otherwise output "NO".

Sample Input

2

3 1 4 1 1 1 1

3 1 4 4 1 4 1

Sample Output

NO

YES

贪心判断就好了。

#include<cstdio>
#include<string>
#include<queue>
#include<cstring>
#include<iostream>
#include<map>
#include<stack>
#include<malloc.h>
#include<algorithm>
using namespace std;
const int maxn = 10000005;
const long long base = 4294967296;
int n, F, T;
long long l, r, a, b, c, d;

void check()
{
    long long y = -1, y0 = -1;
    for (long long i = 1, L = l, R = r, u, v; i <= n; i++)
    {
        u = L; v = R;
        if (u > v) swap(u, v);

        if (v < y || y<0) y = v;

        if (u > y) if (v< y0 || y0 < 0) y0 = v;

        if (u > y0 && y0>0) { F = 1; return; }

        L = (L * a + b) % base;
        R = (R * c + d) % base;
    }
    for (long long i = 1, L = l, R = r, u, v; i <= n; i++)
    {
        u = L; v = R;
        if (u > v) swap(u, v);

        if (v < y || y<0) y = v;

        if (u > y) if (v< y0 || y0 < 0) y0 = v;

        if (u > y0 && y0>0) { F = 1; return; }

        L = (L * a + b) % base;
        R = (R * c + d) % base;
    }
    F = 0;
}

int main()
{
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d", &n);
        F = 0;
        cin >> l >> r >> a >> b >> c >> d;
        check();
        if (F) printf("YES\n"); else printf("NO\n");
    }
    return 0;
}