原题链接 神仙\(DP\)啊... 题解请移步隔壁大佬的博客\(QAQ\) #include<cstdio> using namespace std; const int N = 2e5 + 10; int L[N], R[N], q[N], f[N]; inline int re() { int x = 0; char c = getchar(); bool p = 0; for (; c < '0' || c > '9'; c = getchar()) p |= c == '-…
P3608 [USACO17JAN]Balanced Photo平衡的照片 题目描述 Farmer John is arranging his NN cows in a line to take a photo (1 \leq N \leq 100,0001≤N≤100,000). The height of the iith cow in sequence is h_ihi, and the heights of all cows are distinct. As with all ph…
题目描述 Farmer John has decided to assemble a panoramic photo of a lineup of his N cows (1 <= N <= 200,000), which, as always, are conveniently numbered from 1..N. Accordingly, he snapped M (1 <= M <= 100,000) photos, each covering a contiguous r…
题目链接:传送门 题目: 题目描述 Farmer John has decided to assemble a panoramic photo of a lineup of his N cows ( <= N <= ,), which, ..N. Accordingly, he snapped M ( <= M <= ,) photos, each covering a contiguous range of cows: photo i contains cows a_i thro…
题目描述 一共有n(n≤20000)个人(以1--n编号)向佳佳要照片,而佳佳只能把照片给其中的k个人.佳佳按照与他们的关系好坏的程度给每个人赋予了一个初始权值W[i].然后将初始权值从大到小进行排序,每人就有了一个序号D[i](取值同样是1--n).按照这个序号对10取模的值将这些人分为10类.也就是说定义每个人的类别序号C[i]的值为(D[i]-1) mod 10 +1,显然类别序号的取值为1--10.第i类的人将会额外得到E[i]的权值.你需要做的就是求出加上额外权值以后,最终的权值最大的…
一共有n(n≤20000)个人(以1--n编号)向佳佳要照片,而佳佳只能把照片给其中的k个人.佳佳按照与他们的关系好坏的程度给每个人赋予了一个初始权值W[i].然后将初始权值从大到小进行排序,每人就有了一个序号D[i](取值同样是1--n).按照这个序号对10取模的值将这些人分为10类.也就是说定义每个人的类别序号C[i]的值为(D[i]-1) mod 10 +1,显然类别序号的取值为1--10.第i类的人将会额外得到E[i]的权值.你需要做的就是求出加上额外权值以后,最终的权值最大的k个人,并…
P2209 [USACO13OPEN]燃油经济性Fuel Economy 题目描述 Farmer John has decided to take a cross-country vacation. Not wanting his cows to feel left out, however, he has decided to rent a large truck and to bring the cows with him as well! The truck has a large tan…
https://www.luogu.org/problem/P1583 话不多说,其实就是模拟,然后,各种繁琐 #include<bits/stdc++.h> using namespace std; struct st { int bianhao; int w; int d; int c; int zong; } stu[]; bool cmp1(st a,st b) { if(a.w==b.w) return a.bianhao<b.bianhao; return a.w>b.…
import java.util.*; class Main{ public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); int k = in.nextInt(); int[] extra = new int[11]; for(int i = 1; i <= 10; i ++) { extra[i] = in.nextInt(); } Person[] p…
