1028 : Xiaoming Orders Takeaway

时间限制Time Limit: 1 秒Sec
内存限制Memory Limit: 128 兆MB
提交次数Submitted: 113 次Times
通过次数Solved: 29 次Times

标准评测 Standard

从标准输入读入，结果写到标准输出。评测将输出拆成 token，与标准答案逐项比较，不按整段逐字节比对。 Read from standard input and write to standard output. The judge splits the output into tokens and compares them with the official answer; it does not compare raw bytes.

比对Compare	token 相同即通过。中间的空格、制表符、换行可多可少。Matching tokens pass. Extra spaces, tabs, or newlines between them are ignored.
不同则错Differs	token 个数或内容不同即错误。23 与 2 3、02 与 2、2.0 与 2 均视为不同。A different token count or value is wrong. 23 vs 2 3, 02 vs 2, and 2.0 vs 2 all differ.
输入Input	标准输入，格式见题面。Standard input; format as in the statement.
输出Output	标准输出。Standard output.
时间与内存Time and memory	记这一份程序的 CPU 时间和内存峰值。题目上的时间限制、内存限制就是这一份的上限。The record is this program’s CPU time and peak memory. The limits on the problem apply to this one run.

模板Template

C++ C Java Python Kotlin

#include <iostream>
using namespace std;
int main() {
    int a, b;
    cin >> a >> b;
    cout << (a + b) << endl;
    return 0;
}

#include <stdio.h>
int main() {
    int a, b;
    scanf("%d%d", &a, &b);
    printf("%d\n", a + b);
    return 0;
}

import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int a = in.nextInt();
        int b = in.nextInt();
        System.out.println(a + b);
    }
}

a, b = map(int, input().split())
print(a + b)

import java.util.Scanner
fun main() {
    val inp = Scanner(System.`in`)
    val a = inp.nextInt()
    val b = inp.nextInt()
    println(a + b)
}

题目描述Description

Xiaoming likes to stay at home and takeaways are his favorate. He has accumulated n full discount coupons and wants to make an ordering plan for the next few days.

He will order a total of m takeaways on different days and wants to know the least money will be spent by using coupons wisely.

One order at a time, and at most one coupon can be used for each order.

输入格式Input

There are no more than 100 test cases.

On the first line of each test case, two integer numbers n and m are given. The number of coupons and orders.

Then n lines follow. Each line contains Ci and Di means you need to spend at least Ci to get Di price reduction when using the i-th coupon.

After that, m lines describe the price of the m takeaways. Each line contains Oi means the prise of the i-th order.

1 <= n <= 100, 1 <= m <= 100, 1 <= Di <= Ci <= 100, 1 <= Oi <= 100.

输出格式Output

For each test case output one line, the minimal cost for the orders.

样例Sample

出题Author

CSGrandeur

来源Source

深圳技术大学第一届程序设计竞赛（SZTUCPC2021）

题目到此结束End of problem