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将优惠券的ai进行排序,枚举n张优惠券,若原有的钱m+满减的总金额(bi的和)大于当前ai,则优惠券能用,最后答案为(原有的钱m+(bi之和))的最大值#include <bits/stdc++.h>using namespace std;typedef long long ll;ll t;int n, m;struct discount{ ll re; ll cu;}d[100009];bool cmp(discount a,...
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