SPOJ Solution (in c++)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 //NAME:- ANUHAR TRIPATHI //COLLEGE:- JAYPEE UNIVERSITY OF ENGINEERING & TECHNOLOGY, GUNA //EMAIL:- anuhartripathi15@gmail.com //SPOJ Problem:- AMR10G - Christmas Play # include < bits / stdc ++ .h > # define ll long long using namespace std; int main () { ll t; cin >> t; while (t -- ) { ll n,k; cin >> n >> k; ll h[n],i,j,mi = LONG_MAX; for (i = 0 ;i < n;i ++ ) cin >> h[i]; sort(h,h + n); j = 0 ; for (i = k - 1 ;i < n;i ++ ) { if (mi > h[i] - h[j]) mi = h[i] - h[j]; ++ j; } cout << mi << " \n " ; } return 0 ; }

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 //NAME:- ANUHAR TRIPATHI //COLLEGE:- JAYPEE UNIVERSITY OF ENGINEERING & TECHNOLOGY, GUNA //EMAIL:- anuhartripathi15@gmail.com //SPOJ Problem:- D-query # include < bits / stdc ++ .h > # define ll long long # define bb 555 using namespace std; class node { public: ll l,r,i; }; bool cmp (node x, node y) { if (x.l / bb != y.l / bb) return x.l / bb < y.l / bb; //else if(x.l==y.l) return x.r < y.r; } int main () { ll n,m; cin >> n; ll a[n],i; for (i = 0 ;i < n;i ++ )cin >> a[i]; //memset(a,0,sizeof(a)/sizeof(a[0])); cin >> m; n

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 //NAME:- ANUHAR TRIPATHI //COLLEGE:- JAYPEE UNIVERSITY OF ENGINEERING & TECHNOLOGY, GUNA //EMAIL:- anuhartripathi15@gmail.com //SPOJ Problem:- GSS3 - Can you answer these queries III # include < bits / stdc ++ .h > # define ll long long using namespace std; class node { public: ll suf,pre,result,sum; }tree[ 131080 ]; ll a[ 131080 ]; void build (ll l, ll r, ll pos) { if (l == r) { tree[pos].sum = tree[pos].pre = tree[pos].suf = tree[pos].result = a[l]; return ; } ll mid = (l + r) / 2 ; build(l,mid,pos * 2 ); build(mid + 1 ,r,po