标签：fir clu tput yar osi follow less bsp imu
描述
JAVAMAN is visiting Dream City and he sees a yard of gold coin trees. There are n trees in the yard. Let‘s call them tree 1, tree 2 ...and tree n. At the first day, each tree i has ai coins on it (i=1, 2, 3...n). Surprisingly, each tree i can grow bi new coins each day if it is not cut down. From the first day, JAVAMAN can choose to cut down one tree each day to get all the coins on it. Since he can stay in the Dream City for at most m days, he can cut down at most m trees in all and if he decides not to cut one day, he cannot cut any trees later. (In other words, he can only cut down trees for consecutive m or less days from the first day!)
Given n, m, ai and bi (i=1, 2, 3...n), calculate the maximum number of gold coins JAVAMAN can get.
输入
There are multiple test cases. The first line of input contains an integer T (T <= 200) indicates the number of test cases. Then T test cases follow.
Each test case contains 3 lines: The first line of each test case contains 2 positive integers n and m (0 < m <= n <= 250) separated by a space. The second line of each test case contains n positive integers separated by a space, indicating ai. (0 < ai <= 100, i=1, 2, 3...n) The third line of each test case also contains n positive integers separated by a space, indicating bi. (0 < bi <= 100, i=1, 2, 3...n)
输出
For each test case, output the result in a single line.
样例输入
2
2 1
10 10
1 1
2 2
8 10
2 3
样例输出
10
21
提示
#include <bits/stdc++.h> using namespace std; struct p { int a,b; }x[255]; int dp[255]; bool cmp(p a,p b) { if(a.b!=b.b) return a.b<b.b; return a.a<b.a; } int main() { ios::sync_with_stdio(false); int T; cin>>T; while(T--) { memset(dp,0,sizeof dp); int n,m; cin>>n>>m; for(int i=0;i<n;i++) cin>>x[i].a; for(int i=0;i<n;i++) cin>>x[i].b; sort(x,x+n,cmp); for(int i=0;i<n;i++) for(int j=m;j>0;j--) dp[j]=max(dp[j],dp[j-1]+x[i].a+x[i].b*(j-1)); cout<<dp[m]<<‘\n‘; } return 0; }
标签：fir clu tput yar osi follow less bsp imu
原文地址：https://www.cnblogs.com/zdragon1104/p/9189171.html
