题意描述
John never knew he had a grand-uncle, until he received the notary’s letter. He learned that his late grand-uncle had gathered a lot of money, somewhere in South-America, and that John was the only inheritor.
John did not need that much money for the moment. But he realized that it would be a good idea to store this capital in a safe place, and have it grow until he decided to retire. The bank convinced him that a certain kind of bond was interesting for him.
This kind of bond has a fixed value, and gives a fixed amount of yearly interest, payed to the owner at the end of each year. The bond has no fixed term. Bonds are available in different sizes. The larger ones usually give a better interest. Soon John realized that the optimal set of bonds to buy was not trivial to figure out. Moreover, after a few years his capital would have grown, and the schedule had to be re-evaluated.
Assume the following bonds are available:
Value Annual interest
4000 400
3000 250
With a capital of 10 000 one could buy two bonds of 4 000, giving a yearly interest of 800. Buying two bonds of 3 000, and one of 4 000 is a better idea, as it gives a yearly interest of 900. After two years the capital has grown to 11 800 , and it makes sense to sell a 3 000 one and buy a 4 000 one, so the annual interest grows to 1 050. This is where this story grows unlikely: the bank does not charge for buying and selling bonds. Next year the total sum is 12 850, which allows for three times 4 000, giving a yearly interest of 1 200.
Here is your problem: given an amount to begin with, a number of years, and a set of bonds with their values and interests, find out how big the amount may grow in the given period, using the best schedule for buying and selling bonds.
思路
由于题目给的钱数是1000的倍数，所以我们可以/1000来进行优化。dp[i]表示i*1000块时能够获得的最大利息，然后依次按年来枚举，每次都套一下完全背包模板即可。
AC代码
#include<iostream>
#include<string>
#include<cstring>
#define x first
#define y second
#define PB push_back
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) begin(a),end(a)
#define rep(x,l,u) for(ll x=l;x<u;x++)
#define rrep(x,l,u) for(ll x=l;x>=u;x--)
#define IOS ios::sync_with_stdio(false);cin.tie(0);
using namespace std;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<long,long> PLL;
typedef pair<char,char> PCC;
typedef long long ll;
const int N=15;
const int M=1e6+10;
const int INF=0x3f3f3f3f;
int v[N],w[N],dp[M];
void solve(){
mst(dp,0);
int n,m;cin>>n>>m;
int d;cin>>d;
rep(i,1,d+1){
cin>>v[i]>>w[i];
v[i]/=1000;
}
int ans=n;
rep(i,1,m+1){
int tem=ans/1000;
rep(j,1,d+1){
rep(k,v[j],tem+1){
dp[k]=max(dp[k],dp[k-v[j]]+w[j]);
}
}
ans+=dp[tem];
}
cout<<ans<<endl;
}
int main(){
IOS;
int t;cin>>t;
while(t--){
solve();
}
return 0;
}