variation of coin change DP.

10306.cpp

#include <cstdio>
#include <numeric>
#include <algorithm>
#include <cstring>
using namespace std;
int T, m, S, S2;
int ctype[40][2];
int memo[300][300];
//smallest amount of e-coins needed
int way(int a, int b){
	if((a*a+b*b)>S2) return numeric_limits<int>::max();
	if((a*a+b*b)==S2) return 0;
	if(memo[a][b]!=-1) return memo[a][b];
	int w=numeric_limits<int>::max();
	for(int i=0;i<m;++i) w = min(w, way(a+ctype[i][0], b+ctype[i][1]));
	return memo[a][b] = (w==numeric_limits<int>::max()?numeric_limits<int>::max():1+w);
}

int main(){
	scanf("%d", &T);
	for(int i=0;i<T;++i){
		scanf("%d %d", &m, &S); S2=S*S;
		for(int j=0;j<m;++j){
			scanf("%d %d", &ctype[j][0], &ctype[j][1]);
		}
		memset(memo, -1, sizeof memo);
		int ans=way(0, 0);
		if(ans==numeric_limits<int>::max()) printf("not possible\n");
		else printf("%d\n", ans);
	}
	return 0;
}

passing only one parameter "value" as sum of current chosen coins is not enough since (3,0) and (4,0) can have square sum of 25 but it's not a valid answer.
Thus, we have to track both conventional and InfoTechnological values.

p.s. To differentiate the "not possible" case, we set the minimum answer to INT_MAX. And if after backtracking, the best we got is still INT_MAX, we concluded that there is no way to reach the desired value.