ACM模版

描述

題解

貪心，因為只能一次購買，所以要麼全部按原價求和然後算滿減，要麼能特價的全部算特價，取小的值即可。

程式碼

#include <iostream>

using namespace std;

const int MAXN = 15;

int n, m;
int a[MAXN], b[MAXN];
int c[MAXN], w[MAXN];

int main(int argc, const char * argv[])
{
    while (cin >> n >> m)
    {
        double
 
 sum = 0, ans1 = 0;
        for (int i = 1; i <= n; i++)
        {
            cin >> a[i] >> b[i];
            sum += a[i];
            if (b[i])
            {
                ans1 += a[i] * 0.8;
            }
            else
            {
                ans1 += a[i];
            }
        }

        double
 
 ans2 = sum;
        for (int i = 1; i <= m; i++)
        {
            cin >> c[i] >> w[i];
            if (sum >= c[i] && ans2 > sum - w[i])
            {
                ans2 = sum - w[i];
            }
        }

        if (ans1 > ans2)
        {
            printf
 
("%.2f\n", ans2);
        }
        else
        {
            printf("%.2f\n", ans1);
        }
    }

    return 0;
}