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Time Limit: 20 Sec
Memory Limit: 256 MB
http://acm.hdu.edu.cn/showproblem.php?pid=5265
Input
Output
对于每组的每个询问,输出一行,表示pog与szh的最大恩爱值。
Sample Input
4 4
1 2 3 0
4 4
0 0 2 2
3
2
题意
题解:
STL大法好!
STL拯救世界!
代码:
//qscqesze #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define test freopen("test.txt","r",stdin) #define maxn 2000001 #define mod 10007 #define eps 1e-9 int Num; char CH[20]; const int inf=0x3f3f3f3f; const ll infll = 0x3f3f3f3f3f3f3f3fLL; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} return x*f; } inline void P(int x) { Num=0;if(!x){putchar(‘0‘);puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts(""); } //************************************************************************************** ll a[maxn]; multiset<ll> S; int main() { //test; int n; ll p; while(cin>>n>>p) { S.clear(); for(int i=0;i<n;i++) { a[i]=read(); a[i]%=p; S.insert(a[i]); } ll ans=0; for(int i=0;i<n;i++) { S.erase(S.find(a[i])); multiset<ll>::iterator it=S.lower_bound(p-a[i]); it--; ans=max(ans,(a[i]+*it)%p); it=S.lower_bound(p*2-a[i]); it--; ans=max(ans,(a[i]+*it)%p); S.insert(a[i]); } cout<<ans<<endl; } }
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原文地址:http://www.cnblogs.com/qscqesze/p/4562009.html