Race code "Bestcoder" Cup Chinese college students program Design championship 1001--movie

Problem Description

Cloud and miceren like watching movies.

Today, they want to choose some wonderful scenes from a movie. A movie hasNScenes can chosen, and each scene was associate with an interval [L,< Span id= "mathjax-span-8" class= "Mrow" >r ].  L  is The beginning time of the scene And  r  is the ending time. However, they can ' t choose and 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 so any pair of them does not overlap?

Since There is so many scenes that's can ' t get them in time, we'll give you seven parametersN, l 1, r1 , a, b,  c, d , and you can generateL1 ~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 integersN, l 1, r1 , a, b,  c, d , meaning that there isNScenes. The i-th scene ' s interval is [Li, Ri ].L1 andR1 has been stated in input, andLI=(LI−1∗A+B)MOD 4294967296,RI=(Ri−1 ∗ c + d) mod 4294967296 .

After all the intervals is generated, swap the i-th interval ' sLi andri   If  li > ri .

T is about 100.

1 ≤ N ≤ 10000000.

1 ≤  l1,< Span id= "mathjax-span-219" class= "Msubsup" >r1 ≤ 2000000000 .

1 ≤  a,b,< Span id= "mathjax-span-236" class= "Mi" >c d ≤  1000000000 .

The ratio of test cases with N > are less than 5%.

Output

For each test, print one line.

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

Sample Input

23 1 4 1 1 1 13 1 4 4 1 4 1

Sample Output

NOYES
The main idea: flood. Ask if there are three intervals to meet non-overlapping

#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;}

　　

Race code "Bestcoder" Cup Chinese college students program Design championship 1001--movie