7.24-Codeforces Round #494 (Div. 3)

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Codeforces Round #494 (Div 3) (A~E)

目录 Codeforces 1003 A.Polycarp's Pockets B.Binary String Constructing C.Intense Heat D.Coins and Queries E.Tree Constructing(贪心) F.Abbreviation Codeforces 1003 比赛链接这么做是不是不太好啊233 A.Polycarp's Pockets #include <cstdio> #include <cctype> #include…

Codeforces Round #494 (Div. 3)

刚好在考完当天有一场div3,就开了个小号打了,打的途中被辅导员喊去帮忙,搞了二十分钟-_-||,最后就出了四题,题解如下:题目链接:http://codeforces.com/contest/1003 题目: 思路:求众数出现的次数代码如下: #include <bits/stdc++.h> using namespace std; typedef long long ll; #define debug(x) cout <<"[" <<x <…

Codeforces Round #494 (Div. 3) D. Coins and Queries（贪心

题目链接题目大意:给你n个物品,第iii个物品价值aia_iai,询问q次,问你能不能凑出价值为qiq_iqi的物品. 小贪心吧.从大到小找,能拿就拿就行了. #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mp make_pair #define pb push_back using namespace std; LL gcd(LL a,LL b){retur…

Codeforces Round #494 (Div. 3) D. Coins and Queries (贪心,数学)

题意:给你一组全是\(2^d\ (d\ge0)\)的数,询问q次,每次询问一个数,问这个数是否能够由原数组中的数相加得到,如果能,输出最少用多少个数,否则输出\(-1\). 题解:首先贪心得出结论:如果情况成立,那么最少的情况一定是优先用数组中大的数,然后我们用桶记录数组数的个数,从\(inf\)开始枚举,\(k\)表示桶中的数和\(x\)所需次数的最小值,最后如果\(x\ne 0\)那么条件不满足,否则输出\(ans\)即可. 代码: #include <iostream> #include…

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