uva11386 Triples

这道题其实就是个水题，可惜我忘了一种传说中叫双指针的东西，太弱了，呜呜。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <map>
using namespace std;
int n;
long long a[5010],b[5010];
int num[5010];
template<class T> inline void readint(T& x) {
      char c;
      int mul = 1;
      while((c = getchar()) != EOF){
         if(c == '-')mul = -1;
         if(c >= '0' && c <= '9'){
              x = c-'0';
              break;
         }
      }
      if(c == EOF){x = EOF;return;}
      while(c = getchar()){
          if(c >= '0' && c <= '9'){
              x = (x<<1)+(x<<3)+(c-'0');
          }else break;
      }
      x *= mul;
}
int main(){
    //freopen("in.txt","r",stdin);
    while(~scanf("%d",&n)){
        memset(num,0,sizeof num);
        for(int i=0;i<n;i++) readint(a[i]),b[i]=a[i];
        sort(a,a+n);
        sort(b,b+n);
        int ct=unique(b,b+n)-b;
        for(int i=0;i<n;i++){
           int x=lower_bound(b,b+ct,a[i])-b;
           num[x]++;
        }
        long long ans=0;
        for(int i=0;i<ct;i++){
           int j=i+1;
           int x=lower_bound(b,b+ct,2*b[i])-b;
           if(x<ct&&b[x]==2*b[i]){
              ans+=num[x]*(long long)num[i]*(num[i]-1)/2;
           }
           int k=lower_bound(b,b+ct,b[i]-b[j])-b;
           for(;j<ct;j++){
              while(k<ct&&b[i]+b[j]>b[k]) k++;
              if(k>=ct) continue;
              if(b[i]+b[j]==b[k]){
                 ans+=(long long)num[i]*num[j]*num[k];
              }
           }
        }
        printf("%lld
",ans);
    }
    return 0;
}