Chocolate Bar(暴力)】的更多相关文章

Chocolate Bar Time limit : 2sec / Memory limit : 256MB Score : 400 points Problem Statement There is a bar of chocolate with a height of H blocks and a width of W blocks. Snuke is dividing this bar into exactly three pieces. He can only cut the bar a…
E. Chocolate Bar Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/598/problem/E Description You have a rectangular chocolate bar consisting of n × m single squares. You want to eat exactly k squares, so you may need to break…
Chocolate Bar 题意: 有一个n*m(1<= n,m<=30)的矩形巧克力,每次能横向或者是纵向切,且每次切的花费为所切边长的平方,问你最后得到k个单位巧克力( k <= min(n*m,50) )的最小花费是多少? 思路: 数据规模不大,但是贪心不能得到最优解,很自然想到了dp;里面涉及到行的减少和列的减少,在dp[][]表示中必定要以行数和列数作为dp的含义,但是好像这还不够,如果单单只是一个二维的dp[][]那这个表示的是取了(或者还需)几个单位巧克力呢? ==>…
题目链接:http://codeforces.com/contest/598/problem/E E. Chocolate Bar time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You have a rectangular chocolate bar consisting of n × m single squares. Y…
题目链接: E. Chocolate Bar time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You have a rectangular chocolate bar consisting of n × m single squares. You want to eat exactly k squares, so you ma…
E. Chocolate Bar time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You have a rectangular chocolate bar consisting of n × m single squares. You want to eat exactly k squares, so you may need…
区间DP预处理. dp[i][j][k]表示大小为i*j的巧克力块,切出k块的最小代价. #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; ; ; long long dp[maxn][maxn][maxn]; int n,m,k; void f() { ;i<=;i++) { ;j<=;j++) { ;k<…
Jzzhu and Chocolate time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Jzzhu has a big rectangular chocolate bar that consists of n × m unit squares. He wants to cut this bar exactly k times.…
题意:给出a1*b1和a2*b2两块巧克力,每次可以将这四个数中的随意一个数乘以1/2或者2/3,前提是要可以被2或者3整除,要求最小的次数让a1*b1=a2*b2,并求出这四个数最后的大小. 做法:非常显然仅仅跟2跟3有关.所以s1=a1*b1,s2=a2*b2,s1/=gcd(s1,s2),s2/=gcd(s1,s2),然后若s1跟s2的质因子都是2跟3,那么就有解.之后暴力乱搞就好了. #include<map> #include<string> #include<cs…
D. Chocolate time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Polycarpus likes giving presents to Paraskevi. He has bought two chocolate bars, each of them has the shape of a segmented recta…