codeforces #484B# Maximum Value（二分lower_bound）

B. Maximum Value

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a sequence a consisting of n integers. Find the maximum possible value of  (integer remainder of ai divided by aj), where 1 ≤ i, j ≤ n and ai ≥ aj.

Input

The first line contains integer n — the length of the sequence (1 ≤ n ≤ 2·105).

The second line contains n space-separated integers ai (1 ≤ ai ≤ 106).

Output

Print the answer to the problem.

Sample test(s)

input

3
3 4 5

output

2

解题报告：求一个序列中两数相除的最大余数，保证被除数大于除数。

这题我一开始做的时候是枚举除数，然后找到“除数倍数”位置的前一个位置（lower_bound）,求出它的余数，但是发现不加判重就超时了，加了判重跑得也很吃力。后来听说可以用二分法求余数，自己写了一遍后，运行效率大大提高。所以以后枚举的时候要考虑是不是能够二分。

直接枚举：

/********************************/
/*Problem:  codeforces #484B#   */
/*User: 	    shinelin        */
/*Memory: 	    2100 KB         */
/*Time: 	    639 ms          */
/*Language: 	GNU C++         */
/********************************/
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <ctime>
#include <cctype>
#include <cstring>
#include <string>
#include <list>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <set>
#include <algorithm>
#include <cmath>
using namespace std;

#define INF 0x7fffffff
#define LL long long

vector<int> a;

int main()
{
    int x, N;
    scanf("%d", &N);
    for (int i = 0; i < N; i ++)
    {
        scanf("%d", &x);
        a.push_back(x);
    }
    sort(a.begin(), a.end());
    int MaxMod = 0;
    for (int i = N - 1; i >= 0; i --)
    {
        if(a[i] == 0 || a[i] == a[i-1]) continue; //判重
        int tmp = 2 * a[i];
        while(tmp <= a[N-1])
        {
            int k = lower_bound(a.begin(), a.end(), tmp) - a.begin();
            if(k > 0 && a[k-1] % a[i] > MaxMod)
            {
                MaxMod = a[k-1] % a[i];
                if(MaxMod == a[i] - 1)
                    break;
            }
            tmp += a[i];
        }
        if(a[N-1] % a[i] > MaxMod)
        {
            MaxMod = a[N-1] % a[i];
        }
    }
    printf("%d\n", MaxMod);
    return 0;
}

二分求余数：

/********************************/
/*Problem:  codeforces #484B#   */
/*User: 	    shinelin        */
/*Memory: 	    2100 KB         */
/*Time: 	    109 ms          */
/*Language: 	GNU C++         */
/********************************/
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <ctime>
#include <cctype>
#include <cstring>
#include <string>
#include <list>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <vector>
#include <set>
#include <algorithm>
#include <cmath>
using namespace std;

#define INF 0x7fffffff
#define LL long long

vector<int> a;
int N;

bool BigMod(int d)
{
    int k = lower_bound(a.begin(), a.end(), d) - a.begin();
    for (int i = k; i < N; i ++)
    {
        if(a[i] == a[i-1]) continue;
        int j = 2 * a[i];
        while (j <= a[N-1])
        {
            int cur = lower_bound(a.begin(), a.end(), j) - a.begin();
            if(cur > 0 && a[cur-1] % a[i] > d)
            {
                return true;
            }
            j += a[i];
        }
        if(a[N-1] % a[i] > d)
            return true;
    }
    return false;
}

int main()
{
    int x, Min, Max, Mid;
    scanf("%d", &N);
    for (int i = 0; i < N; i ++)
    {
        scanf("%d", &x);
        a.push_back(x);
    }
    sort(a.begin(), a.end());
    Min = 0;
    Max = a[N-1];
    while(Min < Max)
    {
        Mid = (Max + Min) >> 1;
        if(BigMod(Mid))
        {
            Min = Mid + 1;
        }
        else
        {
            Max = Mid;
        }
    }
    printf("%d\n", Min);
    return 0;
}