4397: [Usaco2015 dec]Breed Counting Time Limit: 10 Sec  Memory Limit: 128 MB Description Farmer John's N cows, conveniently numbered 1…N, are all standing in a row (they seem to do so often that it now takes very little prompting from Farmer John to…
bzoj4397[Usaco2015 dec]Breed Counting 题意: 给定一个长度为N的序列,每个位置上的数只可能是1,2,3中的一种.有Q次询问,每次给定两个数a,b,请分别输出区间[a,b]里数字1,2,3的个数.n≤100000,q≤100000. 题解: 裸前缀和. 代码: #include <cstdio> #include <cstring> #include <algorithm> #define inc(i,j,k) for(int i=j…
原题链接https://www.lydsy.com/JudgeOnline/problem.php?id=4397 用线段树维护区间和即可.时间复杂度\(O((N+Q)logN)\). #include<iostream> #include<cstring> #include<cstdio> #define maxn 100010 using namespace std; inline int read(){ register int x(0),f(1); regist…
4390: [Usaco2015 dec]Max Flow Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 177  Solved: 113[Submit][Status][Discuss] Description Farmer John has installed a new system of N−1 pipes to transport milk between the N stalls in his barn (2≤N≤50,000), c…
Description Bessie the cow is a huge fan of card games, which is quite surprising, given her lack of opposable thumbs. Unfortunately, none of the other cows in the herd are good opponents. They are so bad, in fact, that they always play in a complete…
DP #include<cstdio> using namespace std; int T,A,B,F[5000005],G[5000005]; int main(){ scanf("%d%d%d",&T,&A,&B); F[0]=1; for (int i=0; i<=T; i++) F[i+A]|=F[i],F[i+B]|=F[i]; for (int i=0; i<=T; i++) G[i/2]|=F[i]; for (int i…
4397: [Usaco2015 dec]Breed Counting Time Limit: 10 Sec  Memory Limit: 128 MB Submit: 29  Solved: 25 [Submit][Status][Discuss] Description Farmer John's N cows, conveniently numbered 1-N, are all standing in a row (they seem to do so often that it now…
[BZOJ4391][Usaco2015 dec]High Card Low Card(贪心) 题面 BZOJ 题解 预处理前缀后缀的结果,中间找个地方合并就好了. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<cmath> #include<algorithm> #include<set> using na…
[Usaco2015 dec]Max Flow Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 353  Solved: 236[Submit][Status][Discuss] Description Farmer John has installed a new system of N−1 pipes to transport milk between the N stalls in his barn (2≤N≤50,000), conveni…
题意: T,A,B.T是上限.A和B可以随意吃但是不能超过T.有一次将吃的东西/2的机会.然后可以继续吃,不能超过T.问最多可以吃多少. =>我们先处理不能/2可以吃到哪些.然后弄个双指针扫一扫就可以了TAT #include<cstdio> #include<cstring> #include<cctype> #include<algorithm> using namespace std; #define rep(i,s,t) for(int i=s…