Chef and Magical Jars

Chef and Magical Jars - MAGICJAR

Problem Link

#include <iostream>
using namespace std;

int main()
{
int n, t, dish;
long long maxDish;

cin >> t;
while(t--)
{
cin >> n;
maxDish = 0;
while(n--)
{
cin >> dish;
maxDish += (dish - 1);
}

cout << maxDish + 1 << endl;
}
return 0;
}

Comments

GCJ101BB - Picking Up Chicks

Problem Link /* explanation lets solve the problem only for 2 chicken. s[i] = speed of chicken i pos[i] = position of chicken i if s[i] > s[i - 1] then no problem, just check whether both can reach b within time or not. if s[i] < s[i - 1] then there is a chance that i can slow down i - 1. lets say s[i] = 1 m/sec and s[i - 1] = 2 m/sec and time limit is T and point to reach is B. for s[i] pos[i] can be at max B - T. if pos is greater than B-T it can not reach within Tsec. and for s[i - 1] pos[i - 1] can be at max (B-T)*2. if pos[i - 1] > (B-T)*2 it can not reach within Tsec. at T sec i will be at B and i - 1 will also be at B. at T - 1 i will be at B-T-1 and i-1 will be at B-T-2 and so on. as we can see i -1 will always be behind i. so there will not be any collision. if i is pos[i] < B-T then i can reach B before T sec ...

KOPC12A

KOPC12A - K12 - Building Construction #include <iostream> #include <cmath> #define REP(i, n) for(int i = 0; i < n; i++) using namespace std; const int N = 10005; int height[N]; int cost[N]; int n; //finds the total cost for height h long long int findCost(int h) { long long c = 0; //cost REP(i, n) { c += abs(h - height[i]) * cost[i]; } return c; } int ternary_search(int l, int h) { while(l <= h) { if(l == h) break; int mid1 = l + (h - l) / 3; int mid2 = h - (h - l) / 3; if(findCost(mid1) > findCost(mid2)) l = mid1 + 1; else ...

Cheese and Random Toppings

Problem Link #include <iostream> #include <vector> #include <cstring> using namespace std; #define LL long long int LL lucas(LL n,LL r,LL p) { LL ans=1,ncr[p][p]; memset(ncr, 0, sizeof ncr); for (int i = 0; i < p; ++i) { ncr[i][0]=1; } for (int i = 1; i < p; ++i) { for (int j = 1; j <= i; ++j) { ncr[i][j]=(ncr[i-1][j] + ncr[i-1][j-1])%p; } } while(n && r) { ans=(ans * ncr[n%p][r%p])%p; n/=p; r/=p; } return ans; } LL fastExpo(LL a, LL b, LL P) { LL res = 1; if(b==0) return 1; if(b==1) return a; while (b) { if (b & 1) { res = (res * a) % P; } a = (a * a) % P; b = b >> 1; } return res; } int main() { int t; cin>>t; while(t--) { long long int n,r,m,tm,ans=0; cin>>n>>r>>m; tm=m; for (int i = 2; i <= 50; ++i...

SPOJ

Search This Blog