Tag Archives: logic

Codeforces Beta Round #84 (Div. 1) A. Lucky Sum of Digits

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Problem link: http://codeforces.com/problemset/problem/109/A

A minimum number must contain less number of digits as possible. So to find the minimum number whose digit sum is equal to given n, we have to look into the case 1 and if it fails we have to move to case 2.If case 2 also fails then such minimum lucky  number cannot be formed.

Case 1:if the number can be formed using digits 7 alone.

Case 2: if the number can be formed with combination of digits 4 and 7 such that the number is minimum as possible

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <climits>
#include <stdlib.h>
#include <stdio.h>
using namespace std;
#define REP(i,n) for(int i=0; i<n; i++)
#define FOR(i,st,end) for(int i=st;i<end;i++)
#define db(x) cout << (#x) << " = " << x << endl;
#define mp make_pair
#define pb push_back
#define MAX 10000005
typedef long long int ll;
int main(){
int n;
cin>>n;
if(n<4){
cout<<"-1";
return 0;
}
int numSeven=n/7;
int rem=n%7;
int numFour=rem/4;
int toBecomeFour=rem%4;
if(toBecomeFour!=0){
if(numSeven>=toBecomeFour){
numFour+=((7*toBecomeFour)+toBecomeFour)/4;
numSeven-=toBecomeFour;
}
else{
cout<<"-1";
return 0;
}
}
REP(i,numFour)
cout<<"4";
REP(i,numSeven)
cout<<"7";
}
view raw Lucky Sum of Digits.cpp hosted with ❤ by GitHub

Codeforces #252 (Div.2) C. Valera and Tubes

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Problem link: http://codeforces.com/problemset/problem/441/C

The logic is very simple. Test cases might be misleading. Each tube has to be of length>=2.

So first find the points in zig zag path of the table.(i.e move from left to right and then go the next row and move from right to left and vice versa)

Let the first k-1 tubes be of length 2. Assign 2 points from the path for each tube.

Kth tube(last tube) will have the remaining cells in the zig zag path.

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstring>
#include <climits>
#include <stdlib.h>
#include <stdio.h>
using namespace std;
#define REP(i,n) for(int i=0; i<n; i++)
#define FOR(i,st,end) for(int i=st;i<end;i++)
#define db(x) cout << (#x) << " = " << x << endl;
#define mp make_pair
#define pb push_back
typedef long long int ll;
int arr[305][305];
int visited[305];
int n,m,k;
int withinBoundary(int x,int y){
if(x>=0&&x<n&&y>=0&&y<m)
return 1;
else
return 0;
}
int main(){
cin>>n>>m>>k;
vector<pair<int,int> >q;
int dir=1;
int x=0,y=0;
q.pb(mp(0,0));
while(true){
y+=dir;
if(y==m){
dir*=-1;
y=m-1;
x++;
}
if(y==-1){
dir*=-1;
y=0;
x++;
}
if(x==n)
break;
q.pb(mp(x,y));
}
int j=0;
REP(i,k-1){
cout<<"2 ";
cout<<q[j].first+1<<" "<<q[j].second+1<<" "<<q[j+1].first+1<<" "<<q[j+1].second+1<<endl;
j+=2;
}
cout<<(n*m)-(2*(k-1))<<" ";
while(j<q.size()){
cout<<q[j].first+1<<" "<<q[j].second+1<<" ";
j++;
}
return 0;
}
view raw Valera and Tubes.cpp hosted with ❤ by GitHub