GYM101933I - Intergalactic Bidding

题解：不考虑首先显然是个背包，一开始直接用set模拟，然后map存方案，这样会mle。发现物品的体积有的特殊性

only one participant was allowed to make a bid at a time,
each participant was only allowed to make one bid, and
a participant making a bid had to bid at least twice the amount of the highest
bid at the time.

于是，直接对物品排序贪心取最大即可,因为对于种题目给定的s，一定有唯一的方法组成。

#include <bits/stdc++.h>
typedef long long ll;
const int N = 1e3 + 7;
using namespace std;
struct Big {char x[1005]; int len;string s;bool operator == (const Big &A) const {if(A.len != len) return 0;for(int i = 0; i < len; ++i) if(A.x[i] != x[i]) return 0;return 1;}bool operator < (const Big &A) const {if(A.len != len) return len < A.len;for(int i = len-1; i >= 0; --i) if(A.x[i] != x[i]) return x[i] < A.x[i];return 0;}Big operator + (const Big &A) const {Big ans = *this;for(int i = 0; i < A.len; ++i) {ans.x[i] = ans.x[i] + A.x[i];}int mx = max(A.len, len), f = 0;for(int i = 0; i < mx+1; ++i) {if(ans.x[i] >= 10) {ans.x[i]-=10; ++ans.x[i+1];}if(ans.x[i]) ans.len = i+1, f = 1;}if(!f) ans.len = 1;return ans;}Big operator - (const Big &A) const {Big ans = *this; int f = 0;for(int i = 0; i < A.len; ++i) {ans.x[i] = ans.x[i] - A.x[i];}int mx = max(A.len, len);for(int i = 0; i < mx+1; ++i) {if(ans.x[i] < 0) {ans.x[i]+=10; --ans.x[i+1];}if(ans.x[i]) ans.len = i+1, f = 1;}if(!f) ans.len = 1;return ans;}void read() {cin >> s; len = s.size();for(int i = 0; i < len; ++i) x[len-i-1] = s[i]-'0';s.clear();}void write() {for(int i = len-1; i >= 0; --i) printf("%d",x[i]); putchar('\n');}
} s, ZER;int n;
struct node{string nm; Big A;bool operator < (const node & a) const {return A < a.A;}
} a[N];
vector<int> ans;
int main() {ZER.x[0] = 0; ZER.len = 1;scanf("%d",&n);s.read();for(int i = 1; i <= n; ++i) {cin >> a[i].nm; a[i].A.read();}sort(a+1,a+1+n);for(int i = n ; i >= 1; --i) {if( a[i].A < s || s == a[i].A) {ans.push_back(i);s = s - a[i].A;}}if(s == ZER) {printf("%d\n",(int)ans.size());for(int i = 0; i < ans.size(); ++i) {cout << a[ans[i]].nm << '\n';}}else puts("0");return 0;
}