codeforces 484 B Maximum Value (暴力乱搞)
题目链接 题意：给出n个元素的序列，求出最大的a[i]%a[j] (i>=j) 思路：没思路。。。。
Let us iterate over all different _a__j_. Since we need to maximize  , then iterate all integer _x_ (such_x_ divisible by _a__j_) in range from 2_a__j_ to _M_, where _M_ — doubled maximum value of the sequence. For each such _x_ we need to find maximum _a__i_, such _a__i_ < _x_. Limits for numbers allow to do this in time _O_(1) with an array. After that, update answer by value  . Total time complexity is _O_(_nlogn_ + _MlogM_)
题解也没有特别懂。。。感觉和筛法有点类似。
不过学到了一个o(1)时间得到小于x的最大数是多少的做法。
Sort the array and just maintain another array `A` of `10^6` elements where`index i stores element just smaller than i`For example consider sorted array
[2,4,7,11], then
A(0 indexed) will be[-1,-1,-1,2,2,4,4,4,7,7,7,7,11...]
-1 means no element is smaller than i.
见代码中的b数组。
/* ***********************************************
Author :111qqz
Created Time :2016年03月31日 星期四 14时23分39秒
File Name :code/cf/problem/484B.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int N=1E6+7;
7int b[2*N];
8int n ;
9int main()
10{
11 #ifndef ONLINE_JUDGE
12 freopen("code/in.txt","r",stdin);
13 #endif
14 cin>>n;
15 ms(b,-1);//b[i]表示比小于等于a[i]的元素里最大的。
16 for ( int i = 1 ; i <= n ; i++)
17 {
18 int x;
19 cin>>x;
20 b[x] = x;
21 }
22 for ( int i = 1 ; i <2*N ; i++) if (b[i]==-1) b[i] = b[i-1];
23// for ( int i = 1 ; i <= 20 ; i++) cout<<b[i]<<" ";
24 int ans = 0 ;
25 for ( int j = 1 ; j < N ; j++)
26 if (b[j]==j)
27 for ( int i = j-1 ; i < 2*N ; i+=j)
28 if (j<=b[i]&&ans<b[i]%j) ans = b[i] % j ;
cout<<ans<<endl;
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}