题意:如果一个点,则这点的横竖皆被占领,询问矩阵是否全被占领. 思路:将被占领的x,y标记为1,用x表示1 - i的和 如果x轴的差为 x2 - x1 + 1则表示全被占领,y轴同理 #include <cstdio> #include <cstring> #include <algorithm> #include <functional> #include <vector> #include <queue> typedef long…

Conturbatio Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 786    Accepted Submission(s): 358 Problem Description There are many rook on a chessboard, a rook can attack the row and column it be…

Conturbatio Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 232    Accepted Submission(s): 108 Problem Description There are many rook on a chessboard, a rook can attack the row and column it b…

Conturbatio Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5480 Description There are many rook on a chessboard, a rook can attack the row and column it belongs, including its own place. There are also many quer…

http://acm.hdu.edu.cn/showproblem.php?pid=5084 给出矩阵M,求M*M矩阵的r行c列的数,每个查询跟前一个查询的结果有关. 观察该矩阵得知,令ans = M*M,则 ans[x][y] = (n-1-x行的每个值)*(n-1+y列的每个值),即: ans[x][y] = t[y] * t[2*n - 2 - x] +....+ t[y + n - 1]*t[n - 1 - x] 每一对的和为定值2*n-2-x-y,然后就是求每对i+j的前缀和(所有i+…

Description It has been ten years since TJU-ACM established. And in this year all the retired TJU-ACMers want to get together to celebrate the tenth anniversary. Because the retired TJU-ACMers may live in different places around the world, it may be…

区间求和不更新,开个数组记录一下前缀和就可以了 #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; +; int R[maxn],C[maxn]; int sumR[maxn],sumC[maxn]; int main() { int T; scanf("%d",&T); while(T--) { i…

异或操作蒙蔽了我的双眼 以至于没有第一时间想到前缀和与后缀和 水题做的不够多 #include<bits/stdc++.h> #define rep(i,j,k) for(register int i=j;i<=k;i++) #define rrep(i,j,k) for(register int i=j;i>=k;i--) using namespace std; typedef long long ll; const int maxn = 1e5+11; ll xxor[max…

Taking Bus Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1275    Accepted Submission(s): 420 Problem Description Bestland has a very long road and there are n bus station along the road, whic…

大致题意: 去掉一个元素能使这个数列的GCD最大为多少 分析: 我们求一个数列的GCD,是先求前两个元素的GCD,然后将这个GCD值在与下一个元素进行GCD运算.由此可知进行GCD运算的顺序对最终的结果是没有影响的.我们再看看数列的长度范围,小于100000.那我们就枚举去掉的那个元素,那么去掉元素后的数列的GCD即这个元素前面一段数列的GCD再与这个元素后面一段数列的GCD进行GCD运算.所以我们需要两个数组分别记录前缀GCD和后缀GCD,这两个数组都可以通过O(n)算法算出来. #inclu…