- [NOIP2006 提高组] 金明的预算方案
who can help me 0wa
- 2025-6-6 13:55:34 @
#include<iostream>
using namespace std;
struct main_and_sub{
int ap,ai;
}kb[114514];
int n,sum,m,dp[114514];
int main(){
cin>>n>>m;
n/=10;
for(int i=1;i<=m;i++){
int tp,ti,tm;
cin>>tp>>ti>>tm;
tp/=10;
if(tm==0) kb[i].ap+=tp,kb[i].ai+=ti;
else kb[tm].ap+=tp,kb[tm].ai+=ti;
}
for(int i=1;i<=m;i++){
for(int j=n;j>=kb[i].ap;j--){
dp[j]=max(dp[j],dp[j-kb[i].ap]+kb[i].ap*kb[i].ai);
}
}
cout<<dp[n]*10;
return 0;
}
思路就是把主件和副件整一起当一个算
Information
- ID
- 64
- Time
- 1000ms
- Memory
- 256MiB
- Difficulty
- 4
- Tags
- # Submissions
- 56
- Accepted
- 12
- Uploaded By