# Greedy/two points find Bestcoder Round #43 1002 Pog loves Szh II

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`1 /*2 greedy/Two-point lookup: First for ai%=p, then sort, so that you can use binary search in order. The greedy idea is to find a AJ every time to make and for p-1 (if any)3 of course it is possible to have two numbers and more than P, then the value of an is optimal and is compared with an at a time4 Note: You cannot select two identical numbers5 Reflection: The game thought of%p and Sort,lower_bound, but still did not think of this greedy method to ensure maximum value, or the problem does less AH: (6 */7#include <cstdio>8#include <algorithm>9#include <cstring>Ten#include <cmath> One using namespacestd; A  -typedefLong Longll; - Const intMAXN = 1e5 +Ten; the Const intINF =0x3f3f3f3f; - ll A[MAXN]; -  - intMainvoid)//bestcoder Round #43 1002 Pog loves Szh II + { - //freopen ("b.in", "R", stdin); +  A     intN; ll P; at      while(SCANF ("%d%i64d", &n, &p) = =2) -     { -          for(intI=1; i<=n; ++i) {scanf ("%i64d", &a[i]); A[i]%=p;} -Sort (A +1, A +1+n); -  -ll ans =0; in          for(intI=1; i<=n; ++i) -         { to             intpos = Lower_bound (A +1+i, A +1+n, P-a[i])-A; pos--; +             if(POS <= n && pos! = i) ans = max (ans, (a[i] + a[pos])%p); -             if(I! = n) ans = max (ans, (a[i] + a[n])%p); the         } *  \$printf ("%i64d\n", ans);Panax Notoginseng     } -  the     return 0; +}`

Greedy/two points find Bestcoder Round #43 1002 Pog loves Szh II

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