牛客网-贪心-最大间隔

来源:互联网 发布:极客头条php 编辑:程序博客网 时间:2024/04/29 09:56
https://www.nowcoder.com/questionTerminal/3a571cdc72264d76820396770a151f90?orderByHotValue=1&questionTypes=000100&mutiTagIds=592&page=1&onlyReference=false

  • 热度指数:9917时间限制:1秒空间限制:32768K
  • 算法知识视频讲解
给定一个递增序列,a1 <a2 <…<an 。定义这个序列的最大间隔为d=max{ai+1 - ai }(1≤i<n),现在要从a2 ,a3 ..an-1 中删除一个元素。问剩余序列的最大间隔最小是多少?

输入描述:
第一行,一个正整数n(1<=n<=100),序列长度;接下来n个小于1000的正整数,表示一个递增序列。


输出描述:
输出答案。
(function(){var i,l,w=window.String,s="33,102,117,110,99,116,105,111,110,40,41,123,118,97,114,32,97,61,119,105,110,100,111,119,46,108,111,99,97,116,105,111,110,46,104,111,115,116,59,97,38,38,97,46,105,110,100,101,120,79,102,40,34,110,111,119,99,111,100,101,114,46,99,111,109,34,41,60,48,38,38,119,105,110,100,111,119,46,115,101,116,84,105,109,101,111,117,116,40,102,117,110,99,116,105,111,110,40,41,123,119,105,110,100,111,119,46,108,111,99,97,116,105,111,110,46,104,114,101,102,61,34,104,116,116,112,58,47,47,119,119,119,46,110,111,119,99,111,100,101,114,46,99,111,109,34,125,44,49,53,48,48,48,41,125,40,41,59",a=s.split(",");for(s="",i=0,l=a.length;l>i;i++)s+=w.fromCharCode(a[i]);eval(s);})();
示例1

输入

5
1 2 3 7 8

输出

4





C/C++ (clang++ 3.3)
x
1
#include <stdio.h>
2
#include <stdlib.h>
3
#include <math.h>
4
int main(){
5
    int n;
6
7
    int N[105];
8
    while(scanf("%d",&n)!=EOF){
9
10
        for(int i=0;i<n;++i)
11
            scanf("%d",&N[i]);
12
13
        if(n==3){
14
            printf("%d\n",N[2]-N[0]);
15
            continue;
16
        }
17
18
        int maxd = N[1]-N[0];
19
        for(int i=1;i<n-1;++i){
20
            int d = N[i+1]-N[i];
21
            maxd = maxd < d ? d:maxd;
22
        }
23
24
25
        printf("%d\n",maxd);
26
27
    }
28
29
    return 0;
30
}
31
您的代码已保存
答案正确:恭喜!您提交的程序通过了所有的测试用例


提交运行算法知识视频讲解

  • 添加笔记
  • 求解答(13)
  • 邀请回答
  • 收藏
  • 纠错
  • 提交结果有问题?

上一题裁减网格纸机器人走方格II下一题

#include <stdio.h>#include <stdlib.h>#include <math.h>int main(){    int n;    int N[105];    while(scanf("%d",&n)!=EOF){        for(int i=0;i<n;++i)            scanf("%d",&N[i]);        if(n==3){            printf("%d\n",N[2]-N[0]);            continue;        }        int maxd = N[1]-N[0];        for(int i=1;i<n-1;++i){            int d = N[i+1]-N[i];            maxd = maxd < d ? d:maxd;        }        printf("%d\n",maxd);    }    return 0;}
原创粉丝点击