题意大致是有n个苹果,问你最多拿走m个苹果有多少种拿法.题目非常简单,就是求C(n,0)+...+C(n,m)的组合数的和,但是询问足足有1e5个,然后n,m都是1e5的范围,直接暴力的话肯定时间炸到奶奶都不认识了.当时想了好多好多,各种骚操作都想了一遍就是没想到居然是莫队....我用S(n,m)来记录C(n,0)+...+C(n,m)的和作为一个询问的答案 由组合数公式C(n,m) = C(n-1,m-1)+C(n-1,m)可以推的下面的式子 S(n,m) = S(n,m-1) + C(n,m…

Problem B. Harvest of Apples Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 2397    Accepted Submission(s): 934 Problem Description There are n apples on a tree, numbered from 1 to n.Count th…

题目:http://acm.hdu.edu.cn/showproblem.php?pid=5213 Lucky Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 763    Accepted Submission(s): 249 Problem Description WLD is always very lucky.His secret…

题目大意: 给定n m 在n个数中最多选择m个的所有方案 #include <bits/stdc++.h> using namespace std; #define INF 0x3f3f3f3f #define LL long long ; ; /********组合数模板*********/ LL pow_mod(LL a, LL b) { LL res = 1LL; a %= mod; while(b){ ) res = res * a % mod; a = a * a % mod; b…

Boring counting Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 98304/98304 K (Java/Others)Total Submission(s): 2808    Accepted Submission(s): 826 Problem Description In this problem we consider a rooted tree with N vertices. The vertices ar…

题目链接:Problem - 4638 做了两天莫队和分块,留个模板吧. 当插入r的时候,设arr[r]代表r的位置的数字,判断vis[arr[r-1]]和vis[arr[r+1]]是否访问过,如果两个都访问过,那么块的个数-1,如果有一个访问过,块的个数不变,如果都为0,块的个数+1. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include…

Group Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 2483    Accepted Submission(s): 1272 Problem Description There are n men ,every man has an ID(1..n).their ID is unique. Whose ID is i and i-…

NPY and girls Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 790    Accepted Submission(s): 280 Problem Description NPY's girlfriend blew him out!His honey doesn't love him any more!However, he…

题目描述 分析 \(80\) 分的暴力都打出来了还是没有想到莫队 首先对于 \(s[n][m]\) 我们可以很快地由它推到 \(s[n][m+1]\) 和 \(s[n][m-1]\) 即 \(s[n][m+1]=s[n][m]+C_n^{m+1}\) \(s[n][m-1]=s[n][m]-C_n^m\) 然后我们考虑怎么由 \(s[n][m]\) 推到 \(s[n-1][m]\) 和 \(s[n+1][m]\) 其实画出杨辉三角观察性质即可 摘自 \({\color{black}{M}}{\c…

题意及思路:https://blog.csdn.net/tianyizhicheng/article/details/90369491 代码: #include <bits/stdc++.h> #define lowbit(x) (x & (-x)) using namespace std; const int maxn = 30010; int c[maxn * 3], num[maxn], b[maxn * 3], lb[maxn], rb[maxn], a[maxn]; int…