poj 1046 ——Color Me Less】的更多相关文章

题目来源:POJ 1046 Color Me Less 题目大意:每一个颜色由R.G.B三部分组成,D=Math.sqrt(Math.pow((left.red - right.red), 2)+ Math.pow((left.green - right.green), 2)+ Math.pow((left.blue - right.blue), 2)) 表示两个不同颜色的之间的距离(以left和right为例,left和right分别为两种不同的颜色),现给出16组目标颜色,剩下的为待匹配的颜…
提交地址:http://poj.org/problem?id=1046 Color Me Less Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 32987   Accepted: 16037 Description A color reduction is a mapping from a set of discrete colors to a smaller one. The solution to this pro…
Color Me Less Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 33007   Accepted: 16050 Description A color reduction is a mapping from a set of discrete colors to a smaller one. The solution to this problem requires that you perform just…
一.Description A color reduction is a mapping from a set of discrete colors to a smaller one. The solution to this problem requires that you perform just such a mapping in a standard twenty-four bit RGB color space. The input consists of a target set…
Color Me Less Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 30146   Accepted: 14634 Description A color reduction is a mapping from a set of discrete colors to a smaller one. The solution to this problem requires that you perform just…
Color Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7873   Accepted: 2565 Description Beads of N colors are connected together into a circular necklace of N beads (N<=1000000000). Your job is to calculate how many different kinds of th…
Color Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7630   Accepted: 2507 Description Beads of N colors are connected together into a circular necklace of N beads (N<=1000000000). Your job is to calculate how many different kinds of th…
贪心....                    Color a Tree Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6647   Accepted: 2249 Description Bob is very interested in the data structure of a tree. A tree is a directed graph in which a special node is single…
http://poj.org/problem?id=2777 注意:a可能比b大 #include <cstdio> #include <cstring> #include <algorithm> #define maxn 1000010 using namespace std; int T,L,O,a,b,c; char ch; struct node { int l,r,co; }tree[*maxn]; ]; void build(int l,int r,int…
题目:http://poj.org/problem?id=2154 置换的第二道题! 需要优化!式子是ans=∑n^gcd(i,n)/n (i∈1~n),可以枚举gcd=g,则有phi( n/g )个数与n的gcd是g. g是n的约数,成对出现,可以O(sqrt(n))枚举.用不断 /p 的log(n)做法求单个的phi.(不用专门看p是不是质数,此处可以保证一定是质数) 注意pw里的x传进去要先%mod!!!因为它是1e9级别的,一开始(x*=x)%=mod的时候会爆. 如果把n开成long…