CF #283 (Div. 2) A.（屏蔽数组元素）

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题目链接：http://codeforces.com/contest/496/problem/A

解题思路：

n不是很大，所以暴力。每次屏蔽掉a[ i ]中的一个元素，注意头和尾不能屏蔽。屏蔽后当i == j 时做特殊处理，即cnt = a[ i+ 1 ] - a[ i - 1 ]。最后更新最小值即可。

完整代码：

#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cstring>#include <climits>#include <cassert>#include <complex>#include <cstdio>#include <string>#include <vector>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <list>#include <set>#include <map>using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")typedef long long LL;typedef double DB;typedef unsigned uint;typedef unsigned long long uLL;/** Constant List .. **/ //{const int MOD = int(1e9)+7;const int INF = 0x3f3f3f3f;const LL INFF = 0x3f3f3f3f3f3f3f3fLL;const DB EPS = 1e-9;const DB OO = 1e20;const DB PI = acos(-1.0); //M_PI;int a[1000001];int main(){    #ifdef DoubleQ    freopen("in.txt","r",stdin);    #endif    std::ios::sync_with_stdio(false);    std::cin.tie(0);    int n;    while(cin >> n)    {        for(int i = 0 ; i < n ; i ++)            cin >> a[i];        int sum;        int ans = INF;        int cnt;        for(int j = 1 ; j < n - 1 ; j ++)        {            sum = -INF;            cnt = 0;            for(int i = 1 ; i < n ; i++)            {                if(i == j)                {                    cnt = a[i + 1] - a[i - 1];                    i ++;                }                else                    cnt = a[i] - a[i - 1];                if(cnt > sum)                    sum = cnt;            }            if(sum < ans)                ans = sum;        }        cout << ans << endl;    }}

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