uva 11264 - Coin Collector(dp or 贪心）

Problem E
Coin Collector
Input: Standard Input

Output: Standard Output

 

Our dear Sultan is visiting a country where there are n different types of coin. He wants to collect as many different types of coin as you can. Now if he wants to withdraw X amount of money from a Bank, the Bank will give him this money using following algorithm.

 

withdraw(X){

if( X == 0) return;

Let Y be the highest valued coin that does not exceed X.

Give the customer Y valued coin.

withdraw(X-Y);

}

 

Now Sultan can withdraw any amount of money from the Bank. He should maximize the number of different coins that he can collect in a single withdrawal.

 

Input:

 

First line of the input contains T the number of test cases. Each of the test cases starts with n (1≤n≤1000), the number of different types of coin. Next line contains n integers C₁, C₂, ... , C_n the value of each coin type. C₁<C₂<C₃< … <C_n<1000000000. C₁ equals to 1.

 

Output:

 

For each test case output one line denoting the maximum number of coins that Sultan can collect in a single withdrawal. He can withdraw infinite amount of money from the Bank.

 

Sample Input	Sample Output
2 6 1 2 4 8 16 32 6 1 3 6 8 15 20	6 4

 

 

 

 

 

 

 

 

Problemsetter: Abdullah Al Mahmud

Special Thanks To: Mohammad Mahmudur Rahman

dp[i][j]表示前i个中选j个的最小和。或者贪心。

#include <cstdio>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <iostream>
#include <stack>
#include <set>
#include <cstring>
#include <stdlib.h>
#include <cmath>
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 1000+5;
const LL INF = 10000000000000;

int a[maxn];
LL dp[maxn][maxn];

int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        int n;
        scanf("%d", &n);
        for(int i = 0;i < n;i++){
            scanf("%d", &a[i]);
        }
        for(int i = 0;i < n;i++){
            for(int j = 0;j < maxn;j++){
                dp[i][j] = INF;
            }
        }
        dp[0][0] = 0;
        dp[0][1] = 1;
        for(int i = 1;i < n;i++){
            for(int j = 0;j <= i+1;j++){
                dp[i][j] = dp[i-1][j];
                if(dp[i-1][j-1]+a[i] < a[i+1] || i == n-1)
                    dp[i][j] = min(dp[i][j], dp[i-1][j-1]+a[i]);
            }
        }
        for(int i = n;i >= 0;i--){
            if(dp[n-1][i] != INF){
                printf("%d\n", i);
                break;
            }
        }
    }
    return 0;
}