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 by these parameters.
Input
The first line contains a single integer 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 is less than 5%.
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大意：大水。。问是否存在三个区间满足不重叠
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAX = 10000000+10;
const int inf = 0x3f3f3f3f;
struct edge{
unsigned int l,r;
}t[MAX];
int main()
{
int T,n;
unsigned int a, b, c, d;
scanf("%d",&T);
while(T--){
scanf("%d%d%d%d%d%d%d",&n,&t[1].l,&t[1].r,&a,&b,&c,&d);
for(int i = 2; i <= n ; i++){
t[i].l = t[i-1].l*a + b;
t[i].r = t[i-1].r*c + d;
}
for(int i = 1; i <= n ;i++){
if(t[i].l > t[i].r)
swap(t[i].l,t[i].r);
}
unsigned int min1 = 4294967295UL,max1 = 0;
for(int i = 1; i <= n ; i++){
if(t[i].r < min1)
min1 = t[i].r;
if(t[i].l > max1)
max1 = t[i].l;
}
if(min1 > max1){
printf("NO\n");
continue;
}
int flag = 0;
for(int i = 1; i <= n ; i++){
if(t[i].l > min1 && t[i].r < max1){
printf("YES\n");
flag = 1;
continue;
}
}
if(!flag)
printf("NO\n");
}
return 0;
}