Costume Party poj 3663 c++

Costume Party

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11252 Accepted: 4523

Description

It's Halloween! Farmer John is taking the cows to a costume party, but unfortunately he only has one costume. The costume fits precisely two cows with a length ofS (1 ≤ S ≤ 1,000,000). FJ has N cows (2 ≤ N ≤ 20,000) conveniently numbered 1..N; cowi has length L_i (1 ≤ L_i ≤ 1,000,000). Two cows can fit into the costume if the sum of their lengths is no greater than the length of the costume. FJ wants to know how many pairs of two distinct cows will fit into the costume.

Input

* Line 1: Two space-separated integers: N and S
* Lines 2..N+1: Line i+1 contains a single integer: L_i

Output

* Line 1: A single integer representing the number of pairs of cows FJ can choose. Note that the order of the two cows does not matter.

Sample Input

4 63521

Sample Output

4

 

 

 

 

题意：求n个数字中，有多少组的两个数字之和<=s;

 思路：n的范围较小，用暴力枚举就可以，排序一下，从大到小，然后前面的满足后后面的一定满足，比较省时。

 

 

 下面是代码：

#include<cstdio>#include<algorithm>using namespace std;int len[20200];int main(){    int i,j,n,s,count;    while(scanf("%d%d",&n,&s)!=EOF)    {        count=0;        for(i=0;i<n;i++)             scanf("%d",&len[i]);        sort(len,len+n);        for(i=n-1;i>=0;i--)        {            if(len[i]<s)            {               for(j=i-1;j>=0;j--)               {                    if(len[i]+len[j]<=s)                    {                         count+=j+1;  // 只要是前面的加和满足，后面的加和一定满足；                          break;                    }               }            }         }          printf("%d\n",count);       }    return 0;}