https://csacademy.com/contest/round-39/task/seven-segment-display/
可以知道，只有1是无解
而且肯定是选出来的位数约小越好。
位数 = (k + 6) / 7，因为总是可以通过买7来最大化缩小位数
然后枚举每一位选什么，选的时候，需要的是，(k - cost + 6) / 7应该是比之前少一位的。这就是合法的。
选一个最小的合法的数字即可。
#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
int num[22];
int a[232];
vector<int> vc;
bool dfs(int lef, int pre) {
if (!lef) return true;
if (lef == 1) return false;
for (int i = 0; i <= 6; ++i) {
if (lef >= num[a[i]] && ((lef - num[a[i]] + 6) / 7) == pre - 1) {
vc.push_back(a[i]);
if (dfs(lef - num[a[i]], pre - 1)) return true;
vc.pop_back();
}
}
return false;
}
void work() {
a[0] = 0, a[1] = 1, a[2] = 2, a[3] = 4, a[4] = 6, a[5] = 7, a[6] = 8;
num[1] = 2, num[7] = 3, num[4] = 4, num[2] = 5, num[0] = 6, num[8] = 7;
num[6] = 6;
int k;
scanf("%d", &k);
if (k == 1) {
cout << -1 << endl;
return;
}
if (k <= 7) {
for (int i = 0; i <= 6; ++i) {
if (k == num[a[i]]) {
printf("%d\n", a[i]);
return;
}
}
}
int sel = (k + 6) / 7;
for (int i = 1; i <= 6; ++i) {
vc.push_back(a[i]);
if ((k - num[a[i]] + 6) / 7 == sel - 1) {
if (dfs(k - num[a[i]], (k - num[a[i]] + 6) / 7)) {
for (int i = 0; i < vc.size(); ++i) {
printf("%d", vc[i]);
}
return;
}
}
vc.pop_back();
}
printf("-1\n");
}
int main() {
#ifdef local
freopen("data.txt", "r", stdin);
// freopen("data.txt", "w", stdout);
#endif
work();
return 0;
}
View Code
