[AGC025B]RGB Coloring

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[AGC025B]RGB Coloring 题目大意: 有\(n(n\le3\times10^5)\)个格子,每个格子可以选择涂成红色.蓝色.绿色或不涂色,三种颜色分别产生\(a,b,a+b(a,b\le3\times10^5)\)的收益.问有多少种涂色方案使得总收益为\(k(k\le18\times10^{10})\). 思路: 涂绿色就相当于同时涂了红色和蓝色,因此枚举红色出现次数\(i\)和蓝色出现次数\(j\).答案就是\(\displaystyle\sum_{\substack{0\l…

AtCoder Grand Contest 025 B - RGB Coloring

B - RGB Coloring 求ax + by = k (0<=x<=n && 0<=y<=n)的方案数,最后乘上C(n, x)*C(n,y) 代码: #include<bits/stdc++.h> using namespace std; #define fi first #define se second #define pi acos(-1.0) #define LL long long #define mp make_pair #define…

AGC 025 B - RGB Coloring

B - RGB Coloring Time limit : 2sec / Memory limit : 1024MB Score : 700 points Problem Statement Takahashi has a tower which is divided into N layers. Initially, all the layers are uncolored. Takahashi is going to paint some of the layers in red, gree…

【AtCoder】AGC025题解

A - Digits Sum 枚举即可代码 #include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define space putchar('\n') #define enter putchar(' ') //#define ivorysi using namespace std; typedef long long int64; template<cla…

AGC025简要题解

AGC025简要题解 B RGB Coloring 一道简单题,枚举即可. C Interval Game 考虑可以进行的操作只有两种,即左拉和右拉,连续进行两次相同的操作是没有用的. 左拉时肯定会选择右端点尽量小的,右拉选择左端点尽量大的,所以排序之后贪心即可. D Choosing Points 首先证明对于所有\(d\),假设让两个不能同时选的点之间连一条边,那么结果是一张二分图. \(d\)是奇数可以黑白染色,\(d\)是偶数的时候,显然连边的两点在同一个颜色内.那么我们可以只考虑这个颜…

【AGC025B】RGB Color

[AGC025B]RGB Color 题面描述 Link to Atcoder Link to Luogu Takahashi has a tower which is divided into \(N\) layers. Initially, all the layers are uncolored. Takahashi is going to paint some of the layers in red, green or blue to make a beautiful tower. H…

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