hdu 5265 pog loves szh II

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5265

pog loves szh II

Description

Pog and Szh are playing games.There is a sequence with n numbers, Pog will choose a number A from the sequence. Szh will choose an another number named B from the rest in the sequence. Then the score will be $(A+B)$ mod $p.$They hope to get the largest score.And what is the largest score?

Input

Several groups of data (no more than 5 groups,$n \geq 1000$).

For each case:

The following line contains two integers,$n(2 \leq n \leq 100000)，p(1 \leq p \leq 2^{31}-1)$。

The following line contains $n$ integers $a_i(0 \leq a_i \leq 2^{31}-1)$。

Output

For each case,output an integer means the largest score.

Sample Input

4 4

1 2 3 0

4 4

0 0 2 2

Sample Output

3

2

原先用二分写挂了，估计边界没处理好，换了set好歹过了，罪过，罪过。。

 #include<algorithm>
 #include<iostream>
 #include<cstdlib>
 #include<cstring>
 #include<cstdio>
 #include<set>
 using std::max;
 using std::multiset;
 const int Max_N = ;
 typedef unsigned long long ull;
 ull n, p, arr[Max_N];
 void solve() {
     ull res = ;
     multiset<ull> rec;
     for (int i = ; i < n; i++) {
         scanf("%lld", &arr[i]);
         rec.insert(arr[i] %= p);
     }
     multiset<ull>::iterator ite;
     for (int i = ; i < n; i++) {
         rec.erase(rec.find(arr[i]));
         ite = rec.lower_bound(p - arr[i]);
         ull v1 = *--ite;
         ite = rec.lower_bound( * p - arr[i]);
         ull v2 = *--ite;
         res = max(res, max((v1 + arr[i]) % p, (v2 + arr[i]) % p));
         rec.insert(arr[i]);
     }
     printf("%lld\n", res);
 }
 int main() {
 #ifdef LOCAL
     freopen("in.txt", "r", stdin);
     freopen("out.txt", "w+", stdout);
 #endif
     while (~scanf("%lld %lld", &n, &p)) solve();
     return ;
 }