ACM 第八届山东省赛 J company SDUT 3902
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company
Problem Description
There are n kinds of goods in the company, with each of them has a inventory of and
direct unit benefit . Now you find due to price changes, for any goods sold on day i,
if its direct benefit is val, the total benefit would be i⋅val.
Beginning from the first day, you can and must sell only one good per day until you can't or don't want to do so. If you are allowed to leave some goods unsold, what's the max total benefit you can get in the end?
Input
The first line contains an integers n(1≤n≤1000).
The second line contains n integers val1,val2,..,valn(−100≤.≤100).
The third line contains n integers cnt1,cnt2,..,cntn(1≤≤100).
Output
Output an integer in a single line, indicating the max total benefit.
Example Input
4 -1 -100 5 6 1 1 1 2
Example Output
51
Hint
sell goods whose price with order as -1, 5, 6, 6, the total benefit would be -1*1 + 5*2 + 6*3 + 6*4 = 51.
Author
贪心算法 ,只能用一个for循环 并且 用long long int 数据量会非常大
倒着 从后面考虑 叠楼梯
层数代表天数 有前到后依次增加
代码:
#include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include <stdio.h> #include <string> typedef long long ll; using namespace std; ll a[1000000]; ll dp[1000000]; int main() { int k,i,j,n; while(cin>>n) { memset(a,0,sizeof(a)); memset(dp,0,sizeof(dp)); for(i=0;i<n;i++) { cin>>a[i]; } for(i=0,j=0;i<n;i++) { cin>>k; while(k>1) { a[n+j]=a[i]; j++; k--; } } sort(a,a+n+j); int len=n+j; ll ans=0; ll dsum=0; //dp[len-1]=a[len-1]; for(i=len-1;i>=0;i--) { dsum+=a[i]; if(dsum<0) break; dp[i]=dp[i+1]+dsum; if(ans<dp[i]) ans=dp[i]; } cout<<ans<<endl; } return 0; } /* 5 -1 -1 -1 -1 1 1 1 1 1 1 */