hdu 4379 The More The Better

题目描述：

Given an sequence of numbers {X1,
X2, ... , Xn},
where Xk = (A * k + B) % mod. Your task is to find the maximum sub sequence {Y1,
Y2, ... , Ym}
where every pair of (Yi, Yj)
satisfies Yi + Yj <=
L (1 ≤ i < j ≤ m), and every Yi <= L (1 ≤ i ≤ m ).

Now given n, L, A, B and mod, your task is to figure out the maximum m described above. (1 ≤ n ≤ 2*107,
1 ≤ L ≤ 2*109, 1 ≤ A, B, mod ≤ 109)

题解：
题目可以用反证法证明，加入集合中的数大于l/2的数至多有1个，所以这个题做法如下：
对于所有小于等于L/2 的，统统可以放进来，并设放入的那些的最大值为maxn。然后对于没有放入的数，看看是否有一个数x，满足x+maxn<=l,如果有的话，答案加1.

这个题需要注意的是，竟然卡常数！！long long过不了，int又会溢出。所以我们不妨回过来看看这个式子Xk =
(A * k + B) % mod，可以推出xk=(x(k-1)+a)%mod，这样就用加法来求并取模就不会超过unsigned int。
AC代码1903ms才过

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
unsigned int a,b,k,n,l,last;
unsigned int mod;
bool v[21000000];
int main()
{
while(cin>>n>>l>>a>>b>>mod)
{
for(unsigned int i=1;i<=n;i++)
v[i]=false;
unsigned int ans=0;unsigned int max=0,t;
last=b;
for(unsigned int i=1;i<=n;i++)
{
t=(last+a)%mod;
if(t*2<=l)
{
ans++;v[i]=true;
if(t>max)max=t;
}
last=t;
}
last=b;
for(unsigned int i=1;i<=n;i++)
{
t=(last+a)%mod;
if((!v[i])&&t+max<=l){ans++;break;}
last=t;
}
cout<<ans<<endl;
}
return 0;
}