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: 

With a capital of e10 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.

Input

Output

Sample Input

110000 424000 4003000 250

Sample Output

14050

 1 /************************************************************************* 2     > File Name: beibao.cpp 3     > Author: Mercu 4     > Mail: bkackback1993@gmail.com 5     > Created Time: 2014年07月20日 星期日 12时13分54秒 6  ************************************************************************/ 7  8 #include<iostream> 9 #include<string.h>10 using namespace std;11 12 #define MAXN 1000513 14 int w[MAXN],v[MAXN];15 int f[200000];16 17 int main()18 {19     int a,b,c,d,g,k,j,i;20     cin>>a;21     while(a--)22     {23         memset(w,0,sizeof(w));24         memset(v,0,sizeof(v));25         memset(f,0,sizeof(f));26         cin>>b>>c;27         cin>>d;28         for(i = 0;i < d;i++)29         {30             cin>>w[i]>>v[i];31             w[i] /= 1000;32         }33         int m = b;34         for(k = 0;k < c;k++)35         {36             m = b;37             m /= 1000;38             for(i = 0;i < d;i++)39             {40                 for(j = w[i];j <= m;j++)41                 {42                     f[j] = f[j]>(f[j-w[i]] + v[i])?f[j]:(f[j-w[i]] + v[i]);43                 }44             }45             b += f[m];46         }47         cout<<b<<endl;48     }49     return 0;50 }

 完全背包
     for(k = 0;k < c;k++) //年份   2         { 3             m = b; 4             m /= 1000; 5             for(i = 0;i < d;i++) 6             { 7                 for(j = w[i];j <= m;j++) 8                 { 9                     f[j] = f[j]>(f[j-w[i]] + v[i])?f[j]:(f[j-w[i]] + v[i]);10                 }11             }12             b += f[m];13         }
　　　　　　　　cout<<b<<endl;

0-1背包
         for(i = 1;i <= a;i++)2         {3             for(j = b;j >= t[i];j--)4             {5                 if(t[i] <= j)6                 f[j] = max(f[j],f[j - t[i]] + v[i]);7             }8         }9         cout<<f[b]<<endl;

1 f[j]=max{f[j],f[j-k*c]+k*w}（0<=k*c<=v)2 f[j] = f[j]>(f[j-w[i]] + v[i])?f[j]:(f[j-w[i]] + v[i]); 

上面就是完全背包的状态转移方程,

1 f[j] = max(f[j],f[j - t[i]] + v[i]);2 f[i, j] = max( f[i-1, j-Wi] + Pi (j >= Wi), f[i-1, j] )