随笔分类 - 线段树||树状数组||主席树
摘要:poj3667 #include<cstdio> #include<algorithm> #define lid id << 1 #define rid id << 1 | 1 using namespace std; const int mx = 50010; struct tree{ int l
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摘要:所谓主席树呢,就是对原来的数列[1..n]的每一个前缀[1..i](1≤i≤n)建立一棵线段树,线段树的每一个节点存某个前缀[1..i]中属于区间[L..R]的数一共有多少个(比如根节点是[1..n],一共i个数,sum[root] = i;根节点的左儿子是[1..(L+R)/2],若不大于(L+R
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摘要:题目传送门 Mayor's posters Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 79846 Accepted: 22978 Description The citizens of Bytetown, AB, could
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摘要:题目传送门 Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 25116 Accepted Sub
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摘要:#define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1 const int maxn = 55555; int sum[maxn<<2]; void PushUP(int rt) { sum[rt] = sum[rt<<1]
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摘要:题目传送门 A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 142198 Accepted: 44136 Case Time Limit: 2000MS Descri
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摘要:题目传送门 Balanced Lineup Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 64655 Accepted: 30135 Case Time Limit: 2000MS Description For the dai
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摘要:题目:https://www.nowcoder.com/acm/contest/139/J题意:给出n个数,求 [1,L],[R,n]这两个区间不同数的个数其实你只要把区间扩大一倍,就是求 [R,L+n]这个区间了 求区间内不同数的个数解决方法有很多 像用离线树状数组、离线莫队、线段树、主席树等等
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