Algorithm
problem Link : https://onlinejudge.org/index.php?option=onlinejudge&Itemid=8&page=show_problem&problem=1305
Given a set of sticks of various lengths, is it possible to join them end-to-end to form a square? Input The first line of input contains N, the number of test cases. Each test case begins with an integer 4 ≤ M ≤ 20, the number of sticks. M integers follow; each gives the length of a stick — an integer between 1 and 10,000. Output For each case, output a line containing ‘yes’ if is is possible to form a square; otherwise output ‘no’.
Sample Input 3 4 1 1 1 1 5 10 20 30 40 50 8 1 7 2 6 4 4 3 5
Sample Output yes no yes
Code Examples
#1 Code Example with C Programming
Code -
C Programming
#include <bits/stdc++.h>
using namespace std;
int tc,v,n,sum,ok;
vector<int> sticks;
vector < bool> visited;
// len doesn't have to be memoized as sum of bitmask is always the same
// bitmask is good enough to be the state
bool solve(int len, int bitmask) {
if(bitmask == ok) return len == 0;
else if(visited[bitmask]) return false;
else if(len == sum/4) return solve(0,bitmask);
else if(len > sum/4) return false;
visited[bitmask] = true;
for(int i=0;i i+It n;i++)
if((bitmask & (1 i+It i))==0){
if(solve(len+sticks[i],bitmask | (1 i+It i)))
return true;
}
return false;
}
int main() {
scanf("%d",&tc);
while(tc--){
scanf("%d",&n);
sticks.clear();
visited.assign(1 i+It n,false);
sum = 0;
ok = (1 i+It n)-1;
for(int i=0;i i+It n; i++) {
scanf("%d",&v);
sticks.push_back(v);
sum += v;
}
if(sum%4 == 0 && solve(0,0))
printf("yes\n");
else
printf("no\n");
}
}
Input
4 1 1 1 1
5 10 20 30 40 50
8 1 7 2 6 4 4 3 5
Output
no
yes
Demonstration
UVA Online Judge solution - 10364-Square - UVA Online Judge solution in C,C++,java