Algorithm: POJ 2063 investment (DP full backpack)

The main effect of the topic:

There are n kinds of bonds to buy, each price is a (A is a multiple of 1000), the annual interest is B. A person with a total of money tot (tot is a multiple of 1000), ask him in y years, the maximum can have how much money?

Ideas:

Since both tot and a are multiples of 1000, all can be reduced by 1000 times times in calculation, saving memory and time.

According to greedy thinking, every year with a full backpack to find the year's maximum available interest, and then the next year with the interest after the total money to continue to calculate ...

Code:

 #include <iostream> #include <queue> #include <stack> #include <cstdio> #include <cstr ing> #include <cmath> #include <map> #include <set> #include <string> #define MP Make_pai  
R #define SQ (x) ((x) * (x)) typedef int Int64;  
Const double PI = ACOs (-1.0);  
const int INF = 0X3F3F3F3F;  
      
using namespace Std;  
const int MOD = 1000000007;  
const int MAXN = 13;  
      
const int ADD = 1000*100;  
int tot, y,n;  
int C[MAXN], W[MAXN];  
      
      
int f[100030];  
    int main () {int ncase;  
    scanf ("%d", &ncase);  
        while (ncase--) {scanf ("%d%d%d", &tot, &y, &n);      
            for (int i=0; i<n; ++i) {scanf ("%d%d", &c[i], &w[i]);  
        C[i]/= 1000;  
            for (int i=0; i<y; ++i) {for (int j=0; j<=tot/1000; ++j) f[j] = 0; for (int j=0; j<n; ++j) {for (intV=C[J]; v<=tot/1000;  
            ++V) F[v] = max (F[v], f[v-c[j]]+w[j]);  
        } tot + = f[tot/1000];  
    printf ("%d\n", tot);  
return 0; }

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