技巧（5）数对和小于定值

The More The Better

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3360    Accepted Submission(s): 863

Problem Description

Given an sequence of numbers {X₁, X₂, ... , X_n}, where X_k = (A * k + B) % mod. Your task is to find the maximum sub sequence {Y₁, Y₂, ... , Y_m} where every pair of (Y_i, Y_j) satisfies Y_i + Y_j <= L (1 ≤ i < j ≤ m), and every Y_i <= L (1 ≤ i ≤ m ).
Now given n, L, A, B and mod, your task is to figure out the maximum m described above.

 

 

Input

Multiple test cases, process to the end of input. Every test case has a single line. A line of 5 integers: n, L, A, B and mod. (1 ≤ n ≤ 2*10⁷, 1 ≤ L ≤ 2*10⁹, 1 ≤ A, B, mod ≤ 10⁹)

 

 

Output

For each case, output m in one line.

 

 

Sample Input

1 8 2 3 6

5 8 2 3 6

 

 

Sample Output

1

4

 

 

Source

2012 Multi-University Training Contest 8

 

 一对数(x,y)之和 比 L 小; 要么x < L/2 && y < L/2; 要么 x < L/2 && y > L/2; 不会出现 x > L/2 && y > L/2; 所以对于数组x[1~n]:x[] < L/2 , 一定可以。

并且只能有一个大于 L/2 的 数字x[] , 当然 这要保证 这个 x[] 与 其他所有 小于 L/2 的x[] 的和 都 < L , 才可成立。所以 判断 maxn（x[<=L/2]） + minn(x[>L/2]) 是否 <= L 。

#include<cstdio>
#include<algorithm>
#define ll __int64
using namespace std;
int main(){
    ll n,a,b,mod,l;
    while(~scanf("%I64d%I64d%I64d%I64d%I64d",&n,&l,&a,&b,&mod)){
        ll ans = 0;
        ll maxn = 0;
        ll minn = 2e9;
        for(ll i = 1; i <= n; i++){
            ll x = (a*i+b)%mod;
            if((x*2)<= l){
                ans++;
                maxn = max(x,maxn);
            }
            else {
                minn = min(x,minn);
            }
        }
        if(maxn+minn <= l)ans++;
        printf("%I64d\n",ans);
    }
    return 0;
}