CF1256A Payment Without Change 题解】的更多相关文章

OI生涯打的第一场CF比赛,写篇题解纪念一下吧 ------------可以想到先尽量用面值为1的硬币来凑,然后再用面值为n的硬币来补足.先算出用上所有面值为1的硬币还差多少钱,然后判断用面值为n的硬币够不够补足就可以了.计算需要面值为n的硬币的数量的时候,注意是否需要加1的判断.还有一个需要注意的点,有可能用面值为1的硬币就够了,即$b>=s$,有可能会出现没有特判这个情况导致 Wrong answer on test 2 的情况. #include<iostream> #includ…
CF1256A Payment Without Change 洛谷评测传送门 题目描述 You have aa coins of value nn and bb coins of value 11 . You always pay in exact change, so you want to know if there exist such xx and yy that if you take xx ( 0 \le x \le a0≤x≤a ) coins of value nn and yy…
A. Payment Without Change You have a coins of value n and b coins of value 1. You always pay in exact change, so you want to know if there exist such x and y that if you take x (0≤x≤a) coins of value n and y (0≤y≤b) coins of value 1, then the total v…
Problem A: Small change 题解:http://www.cnblogs.com/crazyapple/p/3349469.html Problem B: Scoop water 题解:http://www.cnblogs.com/crazyapple/p/3349478.html Problem D: CX and girls 题解:http://www.cnblogs.com/crazyapple/p/3349480.html Problem F: ZZY and his…
https://codeforces.com/contest/1256 A:Payment Without Change[思维] 题意:给你a个价值n的物品和b个价值1的物品,问是否存在取物方案使得价值为s 题解:min(s/n,a)*n+b>=s?YES:NO #include<iostream> #include<cstdio> #include<cstdlib> #include<cmath> #include<algorithm>…
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贪心基础 贪心(Greedy)常用于解决最优问题,以期通过某种策略获得一系列局部最优解.从而求得整体最优解. 贪心从局部最优角度考虑,只适用于具备无后效性的问题,即某个状态以前的过程不影响以后的状态.紧接下来的状态仅与当前状态有关.和分治.动态规划一样,贪心是一种思路,不是解决某类问题的具体方法. 应用贪心的关键,是甄别问题是否具备无后效性.找到获得局部最优的策略.有的问题比较浅显,例如一道找零钱的题目 LeetCode 860. Lemonade Change: // 860. Lemonad…
传送门 A. Payment Without Change 签到. Code /* * Author: heyuhhh * Created Time: 2019/11/4 21:19:19 */ #include <bits/stdc++.h> #define MP make_pair #define fi first #define se second #define sz(x) (int)(x).size() #define all(x) (x).begin(), (x).end() #d…
题目链接:https://codeforces.com/problemset/problem/1256/A A. Payment Without Change time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You have aa coins of value nn and bb coins of value 11. You a…
CF598:div3解题报告 A: Payment Without Change 思路: 按题意模拟即可. 代码: #include<bits/stdc++.h> using namespace std; typedef long long ll; int main() { ll T; cin >> T; while(T--) { ll a, b, n, s; cin >> a >> b >> n >> s; if(a*n + b &…