Marsha and Bill own a collection of marbles. They want to split the collection among themselves so that both receive an equal share of the marbles. This would be easy if all the marbles had the same value, because then they could just split the collection in half. But unfortunately, some of the marbles are larger, or more beautiful than others. So, Marsha and Bill start by assigning a value, a natural number between one and six, to each marble. Now they want to divide the marbles so that each of them gets the same total value.
Unfortunately, they realize that it might be impossible to divide the marbles in this way (even if the total value of all marbles is even). For example, if there are one marble of value 1, one of value 3 and two of value 4, then they cannot be split into sets of equal value. So, they ask you to write a program that checks whether there is a fair partition of the marbles.
input
0 0 0 2 0 0
1 0 0 0 1 2
1 0 0 0 1 1
33 12 54 78 65 11
0 0 0 0 0 0
output
Collection #1:
Can be divided.
Collection #2:
Can’t be divided.
Collection #3:
Can be divided.
Collection #4:
Can be divided.
code
//Siberian Squirrel
//#include<bits/stdc++.h>
#include<unordered_map>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<string>
#include<cmath>
#define ACM_LOCAL
using namespace std;
typedef long long ll;
const double PI = acos(-1);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;
const int MOD = 3221225473;
const int N = 2e5 + 10;
const int UP = 101;
int a[7], limit;
int f1[N], f2[N];
inline void solve(ll res = 0) {
for (int i = 1; i <= 6; ++i) {
a[i] %= 60;
res += a[i] * i;
}
if (res & 1) {
printf("Can't be divided.\n");
} else {
memset(f1, 0, sizeof f1);
f1[0] = 1;
for (int i = 1; i <= 6; ++i) {
for (int j = 0; j <= res; ++j) {
for (int k = 0; k <= a[i] && j + k * i <= res; ++k) {
f2[j + k * i] += f1[j];
}
}
for (int j = 0; j <= res; ++j) {
f1[j] = f2[j];
f2[j] = 0;
}
}
if (f1[res / 2]) {
printf("Can be divided.\n");
} else printf("Can't be divided.\n");
}
}
int main() {
#ifdef ACM_LOCAL
freopen("input", "r", stdin);
freopen("output", "w", stdout);
#endif
int o = 1, cases = 0;
// scanf("%d", &o);
while(o --) {
int f = 0;
while(scanf("%d%d%d%d%d%d", &a[1], &a[2], &a[3], &a[4], &a[5], &a[6])) {
f = 0;
for(int i = 1; i <= 6; ++ i) f += a[i];
if(f) {
printf("Collection #%d:\n", ++ cases);
solve();
printf("\n");
} else break;
}
}
return 0;
}