hdu 4476 Cut the rope

Cut the rope

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 206    Accepted Submission(s): 99

Problem Description

　　Now we have some ropes. For each rope, you can cut it only once at any place, or you can also do nothing on it. 
　　If we cut these ropes optimally, how many ropes we can get at most with the same length?

 

Input

　　There is an integer T (1 <= T <= 100) in the first line, indicates there are T test cases in total.
　　For each test case, the first integer N (1 <= N <= 100000) indicates there are N ropes. And then there are N integers whose value are all in [1, 100000], indicate the lengths of these ropes.

 

Output

　　For each test case, you should output one integer in one line, indicates that the number of ropes we can get at most with the same length if we cut these ropes optimally.

 

Sample Input

31 12 1 23 2 3 3

 

Sample Output

235
Hint
　　For Sample 1, we can get two ropes with 0.5 unit length.　　For Sample 2, we can get three ropes with 1 unit length.　　For Sample 3, we can get five ropes with 1.5 unit length
.
题意：n条绳子，每条绳子可以剪一次，也可以不剪，问最多能够得到多少条相同长度的绳子
解题思路：能够想到的一种做法是，枚举每条绳子一半的长度，求最大值。但是n到100000，直接暴力枚举会超时。
用s[i]表示小于等于i长的绳子数，如果当前枚举的绳子长度是a/2，则能够获取的最长相同长度绳子数为：总的绳子数n减去比a/2短的绳子数；还应加上a长的绳子数，因为a长的绳子是对半截取，所以会多出一倍，最后求得最大值。
AC代码：
#include <iostream>#include <cstring>#include <cstdio>#define N 100005using namespace std;int main(){    int t,a,n,f[N],s[N];    scanf("%d",&t);    while(t--)    {        scanf("%d",&n);        memset(f,0,sizeof(f));        for(int i=0;i<n;i++)        {            scanf("%d",&a);            f[a]++;        }        s[0]=0;        for(int i=1;i<N;i++)        s[i]=s[i-1]+f[i];        int max=0,t;        for(int i=1;i<N;i++)        {            if(f[i]>0)            {              t=n-s[(i-1)/2]+f[i];            if(t>max)            max=t;            }        }        printf("%d\n",max);    }return 0;}