max auth uac source name min test targe das

Assignment

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 617 Accepted Submission(s): 314


Problem Description Tom owns a company and he is the boss. There are n staffs which are numbered from 1 to n in this company, and every staff has a ability. Now, Tom is going to assign a special task to some staffs who were in the same group. In a group, the difference of the ability of any two staff is less than k, and their numbers are continuous. Tom want to know the number of groups like this.
Input In the first line a number T indicates the number of test cases. Then for each case the first line contain 2 numbers n, k (1<=n<=100000, 0<k<=10^9),indicate the company has n persons, k means the maximum difference between abilities of staff in a group is less than k. The second line contains n integers:a[1],a[2],…,a[n](0<=a[i]<=10^9),indicate the i-th staff’s ability.
Output For each test。output the number of groups.
Sample Input

2
4 2
3 1 2 4
10 5
0 3 4 5 2 1 6 7 8 9

Sample Output

5
28
HintFirst Sample, the satisfied groups include:[1,1]、[2,2]、[3,3]、[4,4] 、[2,3]  

Author FZUACM
Source

field=problem&key=2015+Multi-University+Training+Contest+1&source=1&searchmode=source" style="color:rgb(26,92,200); text-decoration:none">2015 Multi-University Training Contest 1

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>

using namespace std;

int n,k;
int a[100100];

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        __int64 sum = 0;
        scanf("%d%d",&n,&k);
        for(int i=0; i<n; i++)
        {
            scanf("%d",&a[i]);
        }
        int bmax,bmin;
        for(int i=0; i<n; i++)
        {
            int max = a[i],min = a[i];
            bmax = i;
            bmin = i;
            int j;
            for(j=i; j<n; j++)
            {
                if(max<a[j])
                {
                    bmax = j;
                    max = a[j];
                }
                if(min > a[j])
                {
                    bmin = j;
                    min = a[j];
                }
                if(max-min >=k)
                {
                    int k2 = 1;
                    int k1 = 1;
                    if(bmax>bmin)
                    {
                        for(int v=i; v<bmax; v++)
                        {
                            sum += k1++;
                        }
                        for(int v=bmin+1;v<bmax;v++)
                        {
                            sum -= k2++;
                        }
                        i = bmin;
                    }
                    else
                    {
                        for(int v=i; v<bmin; v++)
                        {
                            sum += k1++;
                        }
                        for(int v=bmax+1;v<bmin;v++)
                        {
                            sum -= k2++;
                        }
                        i = bmax;
                    }
                    break;
                }
            }
            if(j == n)
            {
                int pk = 1;
                for(int v=i;v<n;v++)
                {
                    sum += pk++;
                }
                i = n;
                break;
            }
        }
        printf("%I64d\n",sum);
    }
    return 0;
}

HDU 5289 Assignment（多校聯合第一場1002）