HDU-3348-coins

链接:https://vjudge.net/problem/HDU-3348

题意:

给1,5,10,50,100面值的钞票个a张,求得到p的最小张数和最大张数,

不能则输出-1.

思路:

贪心,正常求最小张数,然后讲所有钞票能表示的总和,减去原本求的价值。

求这个的最小张数,剩下的则是原本价值的最大张数。

代码:

#include <iostream>
#include <memory.h>
#include <string>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <algorithm>
#include <map>
#include <queue>
#include <math.h>
#include <cstdio>
using namespace std;

typedef long long LL;

int a[10] = {0, 1, 5, 10, 50, 100};
int v[10];

int Get_M(int value)
{
    int res = 0;
    for (int i = 5;i >= 1;i--)
    {
        if (value >= a[i])
        {
            int tmp = value / a[i];
            value -= a[i] * min(tmp, v[i]);
            res += min(tmp, v[i]);
        }
    }
    if (value != 0)
        return -1;
    return res;
}

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int p;
        int sum = 0, num = 0;
        cin >> p;
        for (int i = 1;i <= 5;i++)
        {
            cin >> v[i];
            num += v[i];
            sum += v[i] * a[i];
        }
        int res_min = 0, res_max = 0;
        res_min = Get_M(p);
        res_max = Get_M(sum - p);
        if (res_min == -1 || res_max == -1)
            cout << -1 << ' ' << -1 << endl;
        else
            cout << res_min << ' ' << num - res_max << endl;
    }

    return 0;
}

  

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转载自www.cnblogs.com/YDDDD/p/10624987.html
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