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Spoj EDIST Solution

on September 24, 2022

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Spoj KNAPSACK Solution

on March 06, 2022

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Spoj INVCNT Solution

on October 20, 2021

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Spoj QTREE2 Solution

on September 21, 2021

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Spoj MULTQ3 Solution

on September 04, 2021

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Spoj ETF Solution

on June 02, 2021

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Spoj DIVSUM2 Solution

on June 01, 2021

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Spoj NUMDIV Solution

on June 01, 2021

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Spoj PON Solution

on June 01, 2021

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Spoj DIVSUM Solution

on May 29, 2021

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Spoj COMDIV Solution

on May 29, 2021

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Spoj CONSEC Solution

on May 23, 2021

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Spoj CLFLARR Solution

on May 22, 2021

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Spoj EZDIJKST Solution

on July 02, 2020

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Spoj SHPATH Solution

on July 02, 2020

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Spoj BUGLIFE Solution

on June 25, 2020

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Uva 12347 Solution

#include<bits/stdc++.h> using namespace std; struct node{ int data; node *left; node *right; }; node *newnode(int data){ node *temp = (node*) malloc(sizeof(node)); temp->data = data; temp->left = temp->right = NULL; return temp; } node *inserti(node *node, int data) { if(node == NULL) return newnode(data); else if(data < node->data) node->left = inserti(node->left, data); else if(data > node->data) node->right = inserti(node->right, data); return node; } void postorder(node *node){ if(node == NULL) return; postorder(node->left); postorder(node->right); cout...

Spoj CLFLARR Solution

#include<bits/stdc++.h> using namespace std; int parent[3000000]; int Find(int x) { if(parent[x] == x) return x; return parent[x] = Find(parent[x]); } void Union(int a, int b) { int u = Find(a); int v = Find(b); int indx = max(u, v); /// max() function use for update parent index parent[u] = indx; parent[v] = indx; } int main() { int n, q; cin >> n >> q; int l[q+5], r[q+5], c[q+5], v[n+5]; for(int i=1;i<=n+1;i++) { parent[i] = i; v[i] = 0; } for(int i=0;i<q;i++) cin >> l[i] >> r[i] >> c[i]; for(int i=q-1;i>=0;i--) { int left = Find(l[i]); while(left <= r[i]) { v[left] = c[i]; Union(left, left+1); /// for index update left = Find(left); /// for index increment } } for(int i=1;i<=n;i++) cout << v[i] << "\n"; return 0; }

URI 1094 Solution

#include<stdio.h> int main() { int an, i, sumc = 0, sumr = 0, sums = 0, sum = 0, n; double a, b, c; char ch; scanf("%d", &n); for(i=0; i<n; i++) { scanf("%d %c",&an,&ch); if(ch == 'C') { sumc = sumc + an; } else if(ch == 'R') { sumr = sumr + an; } else if(ch == 'S') { sums = sums + an; } sum = sum + an; } a = ((float)sumc*100)/(float)sum; b = ((float)sumr*100)/(float)sum; ...

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