2018 Multi-University Training Contest 2: 1007. Naive Operations

题意：给你一个长度为n的b[]数组和长度为n的a[]数组，b[]是1到n的全排列，a[]初始全为0，有两种操作：①add l r：a[]数组从l到r全部+1，②query l r：查询∑a[i]/b[i]（l≤i≤r，向下取整）

因为b[]是n的一个全排列，并且对于所有的a[i]/b[i]很显然只会变化n/b[i]次，所以总共只会有n/1+n/2+…+n/3≈nlogn次变化
可以直接暴力每个位置上的每次变化（从0到n/b[i]）会在第几次add操作之后
可以用二分，如果第k次add操作不会影响当前位置，那么这次操作就相当于0，否则相当于1，这里用前缀和处理，不过是要带修改的，所以可能要用树状数组，看前缀的1的数量s是否满足条件即可
总复杂度O(nlog²n)（好像还会多一个log，不过到不了极端情况）

#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<vector>
using namespace std;
typedef struct
{int l, r;int id, t;
}Que;
Que s[100005];
int len, b[100005], tre[100005];
char str[8];
vector<int> L[100005], R[100005], A[100005];
void Update(int k, int x)
{while(k<=len){tre[k] += x;k += k&-k;}
}
int Query(int k)
{int ans = 0;while(k){ans += tre[k];k -= k&-k;}return ans;
}
int main(void)
{int n, i, j, l, r, m, last, mid, ans;while(scanf("%d%d", &n, &m)!=EOF){for(i=1;i<=n;i++)scanf("%d", &b[i]);len = m;for(i=1;i<=m;i++){scanf("%s%d%d", str+1, &s[i].l, &s[i].r);s[i].id = i;if(str[4]==0){s[i].t = 1;L[s[i].l].push_back(s[i].id);R[s[i].r+1].push_back(s[i].id);}elses[i].t = 0;}for(i=1;i<=n;i++){for(j=0;j<L[i].size();j++)Update(L[i][j], 1);for(j=0;j<R[i].size();j++)Update(R[i][j], -1);last = 1;for(j=1;j<=n/b[i];j++){l = last, r = m+1;while(l<r){mid = (l+r)>>1;if(Query(mid)<j*b[i])l = mid+1;elser = mid;}//printf("%d\n", r);if(r==m+1)break;A[r].push_back(i);last = r+1;}}ans = 0, len = n;memset(tre, 0, sizeof(tre));for(i=1;i<=m;i++){for(j=0;j<A[i].size();j++)Update(A[i][j], 1);if(s[i].t==0)printf("%d\n", Query(s[i].r)-Query(s[i].l-1));A[i].clear();}for(i=1;i<=n;i++){L[i].clear();R[i].clear();tre[i] = 0;}}return 0;
}
/*
5 4
1 5 2 4 3
add 1 4
query 1 4
add 2 5
query 2 5
*/

2018 Multi-University Training Contest 2: 1007. Naive Operations相关推荐

HDU 6356.Glad You Came-线段树(区间更新+剪枝) (2018 Multi-University Training Contest 5 1007)...
6356.Glad You Came 题意就是给你一个随机生成函数,然后从随机函数里确定查询的左右区间以及要更新的val值.然后最后求一下异或和就可以了. 线段树,区间最大值和最小值维护一下,因为数据 ...
2019 Multi-University Training Contest 9 1007 Rikka with Travels
HDU 6686 Rikka with Travels 题意: 在一颗树上选择两条不相交的路径的可能性有多少,路径长度定义为路径的顶点数. 题解: 初步思考,观察样例可以发现,求的是两条路径的有序对, ...
Sichuan University Programming Contest 2018 Preliminary
嗯为了防止大家AK,所以这次的A题和K题我们就当做不存在好了! 经历了昨天写了两个多小时的博客没保存的心态炸裂,今天终于下了个Markdown.所以我猜这篇的格式应该会更好看一点! 好吧废话不多说题 ...
2018 Multi-University Training Contest 3 Problem F. Grab The Tree 【YY+BFS】
传送门:http://acm.hdu.edu.cn/showproblem.php?pid=6324 Problem F. Grab The Tree Time Limit: 2000/1000 MS ...
2018 Multi-University Training Contest 4 Problem E. Matrix from Arrays 【打表+二维前缀和】
任意门:http://acm.hdu.edu.cn/showproblem.php?pid=6336 Problem E. Matrix from Arrays Time Limit: 4000/20 ...
2021 HZNU Winter Training Day 17 (2018 German Collegiate Programming Contest (GCPC 18))
2021 HZNU Winter Training Day 17 (2018 German Collegiate Programming Contest (GCPC 18)) 题目 A B C D E ...
2018 Multi-University Training Contest 7 Age of Moyu
Age of Moyu 来追梦 Problem Description Mr.Quin love fishes so much and Mr.Quin's city has a nautical sy ...
2018 Multi-University Training Contest 2
1003:Cover 巧妙的构造,先找出所有奇数点,再成对连边,在新图里面跑欧拉回路(dfs边到底),切开之前连的边就是划分方案,这样能得到答案是max(d/2,1) (d为奇数点的个数) 1 #in ...
题解：2018 Multi-University Training Contest 2
Game 先手的人可以通过选1,不选1的方式操纵游戏的进程,因此有必胜策略. #include<stdio.h> int main() {for(int n; ~scanf("% ...

2018 Multi-University Training Contest 2: 1007. Naive Operations

2018 Multi-University Training Contest 2: 1007. Naive Operations相关推荐

最新文章

热门文章