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upper_bound和lower_bound用法(STL)

2018-01-22 22:31 465 查看

ABCDEF - ABCDEF

#binary-search

You are given a set S of integers between -30000 and 30000 (inclusive).
Find the total number of sextuples

that satisfy:

Input

The first line contains integer N (1 ≤ N ≤ 100), the size of a set S.
Elements of S are given in the next N lines, one integer per line. Given numbers will be distinct.

Output

Output the total number of plausible sextuples.

Examples

Input:
1
1

Output:
1

Input:
2
2
3

Output:
4

Input:
2
-1
1

Output:
24

Input:
3
5
7
10

Output:
10

#include <iostream>
#include <vector>
#include <algorithm>
#include <sstream>
#include <cstring>
#include <cmath>

using namespace std;

#define MAX_N 101

int n,x[MAX_N],lo,hi;
long long res=0LL;
vector<int>s1,s2;

int main()
{
scanf("%d",&n);
for (int i=0;i<n;i++)
{
scanf("%d",&x[i]);
}
for (int i=0;i<n;i++)
{
for (int j=0;j<n;j++)
{
for (int k=0;k<n;k++)
{
s1.push_back(x[i]*x[j]+x[k]);
}
}
}
for (int i=0;i<n;i++)
{
for (int j=0;j<n;j++)
{
for (int k=0;k<n;k++)
{
if (x[k]==0) continue;
s2.push_back((x[i]+x[j])*x[k]);
}
}
}
sort(s1.begin(),s1.end());
sort(s2.begin(),s2.end());
for (int i=0;i<s1.size();i++)
{
lo=lower_bound(s2.begin(),s2.end(),s1[i])-s2.begin();
hi=upper_bound(s2.begin(),s2.end(),s1[i])-s2.begin();
res+=(hi-lo);
}
printf("%lld\n",res);
return 0;
}

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