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  • pairs

    pairs

    http://acm.hdu.edu.cn/showproblem.php?pid=5178

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 4888    Accepted Submission(s): 1758


    Problem Description
    John has n points on the X axis, and their coordinates are (x[i],0),(i=0,1,2,,n1). He wants to know how many pairs<a,b> that |x[b]x[a]|k.(a<b)
     
    Input
    The first line contains a single integer T (about 5), indicating the number of cases.
    Each test case begins with two integers n,k(1n100000,1k109).
    Next n lines contain an integer x[i](109x[i]109), means the X coordinates.
     
    Output
    For each case, output an integer means how many pairs<a,b> that |x[b]x[a]|k.
     
    Sample Input
    2
    5 5
    -100
    0
    100
    101
    102
    5 300
    -100
    0
    100
    101
    102
     
    Sample Output
    3
    10
     
    Source
     
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    二分水题

     1 #include <cstdio>
     2 #include <cstring>
     3 #include <string>
     4 #include <cmath>
     5 #include <iostream>
     6 #include <algorithm>
     7 #include <queue>
     8 #include <stack>
     9 #include <vector>
    10 #include <set>
    11 #include <map>
    12 typedef long long ll;
    13 #define maxn 100005
    14 #define lson l,mid,rt<<1
    15 #define rson mid+1,r,rt<<1|1
    16 using namespace std;
    17 
    18 
    19 ll n,k;
    20 ll a[maxn];
    21 
    22 int main(){
    23 
    24     int T;
    25     while(cin>>T){
    26         while(T--){
    27             cin>>n>>k;
    28             for(int i=1;i<=n;i++){
    29                 cin>>a[i];
    30             }
    31             sort(a+1,a+n+1);
    32             ll ans=0;
    33             int p;
    34             for(int i=2;i<=n;i++){
    35                 p=lower_bound(a+1,a+i,a[i]-k)-a;
    36                 ans+=i-p;
    37             }
    38 
    39             cout<<ans<<endl;
    40         }
    41     }
    42 
    43 }
    View Code
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  • 原文地址:https://www.cnblogs.com/Fighting-sh/p/10073609.html
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