Time Limit: 1000 ms Memory Limit: 65536 KiB
Submit Statistic
Problem Description
There are n coins with different denominations, and the denomination of each coin is stored in the array T[1:n]. Now I want to use these coins to find money. The number of coins of various denominations that can be used is stored in the array Coins[1:n].
For any amount of money 0≤m≤20001, design a method to find money m with the least number of coins.
For a given 1≤n≤10, the coin denomination array T and the coin number array Coins of various denominations that can be used, and the amount of money m, 0≤m≤20001, calculate Find the minimum number of coins for money m.
Input
In the first line of the input data, there is only 1 integer giving the value of n, and 2 numbers in each line starting from the second line, which are T[j] and Coins[j]. The last line is m.
Output
The output data has only one integer, which represents the minimum number of coins calculated. When the problem has no solution, -1 is output.
Sample Input
3
1 3
2 3
5 3
18
Sample Output
5
Hint

Source

//There are restrictions on the types of coins and the number of coins
//Multiple backpacks

#include <iostream>
//#include <cstdio>
using namespace std;
const int maxvalue = 20001;
const int coinnum = 15;
int mymin(int a, int b)
{
    return a < b ? a : b;
}

int main()
{
    int n;
    cin >> n;
    int coin[coinnum];
    int coincount[coinnum];
    for(int i = 0; i < n; i++)
    {
        cin >> coin[i];
        cin >> coincount[i];
    }
    int m;
    cin >> m;
    //The amount of money is dp[i] is the number of coins
    int *dp = new int[maxvalue]();
    //Key error
    //for(int i = 0; i <= m; i++) is wrong
    for(int i = 1; i <= m;i++)
        dp[i] = maxvalue;
    int i, j, k;
    for(i = 0; i < n; i++)//The number of coin denominations
    {
        for(j = 1; j <= coincount[i]; j++)//The number of denominations of coins
        {
            for(k = m; k >= coin[i]; k--)
            {
                //The dynamic migration equation is
                dp[k] = mymin(dp[k - coin[i]] + 1, dp[k]);
            }
        }
    }
    if(dp[m] == maxvalue)
        cout << -1 << endl;
    else cout << dp[m] << endl;
    return 0;
}