ACM-ICPC 2018 南京赛区网络预赛 B. The writing on the wall (暴力）

【ACM-ICPC 2018 南京赛区网络预赛 B. The writing on the wall (暴力）】的更多相关文章

ACM-ICPC 2018 南京赛区网络预赛 B. The writing on the wall (暴力）

题意:一个n*m的方格矩阵,有的格子被涂成了黑色,问该矩阵中有多少个子矩阵,子矩阵不包含黑色格子; 思路:对于一个长为L, 高为H的无黑点矩阵中包含的高为H的子矩阵个数为L+(L-1)+(L-2)+...+1个:这是直接算的一种方法:如何程序表示该计算呢? ; i<=L; i++){ ; j--){ count+=; } } 这样的一个双层循环就表示了上式:那么所有子矩阵个数就是三层循环,高由1->H: ; h<=H; h++){ ; i<=L; i++){ ; j--){ cou…

ACM-ICPC 2018 南京赛区网络预赛 B. The writing on the wall

题目链接:https://nanti.jisuanke.com/t/30991 2000ms 262144K Feeling hungry, a cute hamster decides to order some take-away food (like fried chicken for only 3030 Yuan). However, his owner CXY thinks that take-away food is unhealthy and expensive. So she…

ACM-ICPC 2018 南京赛区网络预赛 B The writing on the wall(思维)

https://nanti.jisuanke.com/t/30991 题意一个n*m的方格矩阵,有的格子被涂成了黑色,问该矩阵中有多少个子矩阵,子矩阵不包含黑色格子. 分析参考https://blog.csdn.net/Sirius_han/article/details/82313029 n=1e5,m=100.首先思考一下没有黑点的话,子矩阵总数怎么算. 现有长为L的大矩阵,对于固定高度h,其子矩阵的个数是这样计算的 ; i<=L; i++){ ; j--){ count+=h; } }…

ACM-ICPC 2018 南京赛区网络预赛 J.sum

A square-free integer is an integer which is indivisible by any square number except 11. For example, 6 = 2 \cdot 36=2⋅3 is square-free, but 12 = 2^2 \cdot 312=22⋅3 is not, because 2^222 is a square number. Some integers could be decomposed into prod…

ACM-ICPC 2018 南京赛区网络预赛 E题

ACM-ICPC 2018 南京赛区网络预赛 E题题目链接: https://nanti.jisuanke.com/t/30994 Dlsj is competing in a contest with n (0 < n \le 20)n(0<n≤20) problems. And he knows the answer of all of these problems. However, he can submit ii-th problem if and only if he has s…