HZAU 1209 Deadline

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1209: Deadline

Time Limit: 2 Sec Memory Limit: 1280 MB
Submit: 1046 Solved: 102
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Description

There are N bugs to be repaired and some engineers whose abilities are roughly equal. And an engineer can repair a bug per day. Each bug has a deadline A[i].

Question: How many engineers can repair all bugs before those deadlines at least?

1<=n<= 1e6. 1<=a[i] <=1e9

Input

There are multiply test cases.

In each case, the first line is an integer N , indicates the number of bugs. The next line is n integers indicates the deadlines of those bugs.

Output

There are one number indicates the answer to the question in a line for each case.

Sample Input

4 1 2 3 4

Sample Output

HINT

题意：

有n(1<=n<=1000000)个bug需要程序猿来调试，程序员每天可以调试出1个bug，每个bug都必须在xi天前(包括xi)调试出来，求出所需最小程序猿的个数。

思路：

我们首先可以把最后期限xi大于n的不进行统计，因为xi大于n时最多也就只需要1个程序员，那么我们枚举天数1到n一共有多少个bug，bug的数量除于天数可得出到该天时所需要的程序员，如果还有余数那么我们还得再增加一个程序员。

举个例子：

1 2 2 3 4

第1天：1/1=1，num=1

第2天：3/2=1......1，num=2

第3天：4/3=1......1，num=2

第4天：5/4=1......1，num=2

第5天：5/5=1，num=1

所需最少的程序员就是2个。

Source

示例程序

#include <cstdio>#include <cstring>#include <algorithm>int a[1000001];//n最大为1000000，因此数组开到1000000即可using namespace std;int main(){    int n,i,x,maxx,num,t;    while(scanf("%d",&n)!=EOF)    {        memset(a,0,sizeof(a));        maxx=1;        num=0;        for(i=1;n>=i;i++)        {            scanf("%d",&x);            if(x<=n)            {                a[x]++;            }        }        for(i=1;n>=i;i++)        {            num=num+a[i];            t=num/i;            if(num%i!=0)            {                t++;            }            maxx=max(maxx,t);        }        printf("%d\n",maxx);    }    return 0;} /**************************************************************    Problem: 1209    Code Length: 740 B    Language: C++    Result: Accepted    Time:1088 ms    Memory:4940 kb****************************************************************/

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