树状数组.Easy. /* 4267 */ #include <iostream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #include <vector> #include <deque> #include <algorithm> #include <cstdio&g…

A Simple Problem with Integers Time Limit:5000MS   Memory Limit:131072K Case Time Limit:2000MS Description You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each…

这题用线段树轻松解了,重新用树状数组解,关键点是区间更新.公式推导如下:sum[x] = org_sum[x] + delta[1]*x + delta[2]*(x-1) + delta[x]*1           = org_sum[x] + Sigma(delta[1..x]) * (x+1) - Sigma(delta[i]*i)树状数组增加两个结点信息分别存delta[i] 和 delta[i] * i就好了. /* 3468 */ #include <iostream> #incl…

A Simple Problem with Integers Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5339    Accepted Submission(s): 1693 Problem Description Let A1, A2, ... , AN be N elements. You need to deal with…

A Simple Problem with Integers Time Limit: 5000/1500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5402    Accepted Submission(s): 1710 Problem Description Let A1, A2, ... , AN be N elements. You need to deal with…

http://acm.hdu.edu.cn/showproblem.php?pid=4267 [思路] 树状数组的区间修改:在区间[a, b]内更新+x就在a的位置+x. 然后在b+1的位置-x 树状数组的单点查询:求某点a的值就是求数组中1~a的和. (i-a)%k==0把区间分隔开了,不能直接套用树状数组的区间修改单点查询 这道题的K很小,所以可以枚举k,对于每个k,建立k个树状数组,所以一共建立55棵树 所以就可以多建几棵树..然后就可以转换为成段更新了~~ [AC] #include<b…

水题. #include <cstdio> #include <cstring> #include <cstdlib> int abs(int x) { ? -x:x; } int main() { int t, n; int cur, past; __int64 ans; bool flag; #ifndef ONLINE_JUDGE freopen("data.in", "r", stdin); #endif scanf(&q…

其实是求树上的路径间的数据第K大的题目.果断主席树 + LCA.初始流量是这条路径上的最小值.若a<=b,显然直接为s->t建立pipe可以使流量最优:否则,对[0, 10**4]二分得到boundry,使得boundry * n_edge - sum_edge <= k/b, 或者建立s->t,然后不断extend s->t. /* 4729 */ #include <iostream> #include <sstream> #include <…

题目: Problem Description Let A1, A2, ... , AN be N elements. You need to deal with two kinds of operations. One type of operation is to add a given number to a few numbers in a given interval. The other is to query the value of some element.   Input T…

显然需要贪心,重叠越长越好,这样最终的串长尽可能短.需要注意的是,不要考虑中间结果,显然是个状态dp.先做预处理去重,然后求任意一对串的公共长度. /* 3828 */ #include <iostream> #include <sstream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #includ…