1048 Find Coins

最新推荐文章于 2023-09-30 11:45:59 发布

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解决PAT算法题中关于使用两种特定硬币支付的问题，通过数组记录硬币面值并查找匹配项。

Eva loves to collect coins from all over the universe, including some other planets like Mars. One day she visited a universal shopping mall which could accept all kinds of coins as payments. However, there was a special requirement of the payment: for each bill, she could only use exactly two coins to pay the exact amount. Since she has as many as 105 coins with her, she definitely needs your help. You are supposed to tell her, for any given amount of money, whether or not she can find two coins to pay for it.

Input Specification:

Each input file contains one test case. For each case, the first line contains 2 positive numbers: N (<=105, the total number of coins) and M(<=103, the amount of money Eva has to pay). The second line contains N face values of the coins, which are all positive numbers no more than 500. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the two face values V1 and V2 (separated by a space) such that V1 + V2 = M and V1 <= V2. If such a solution is not unique, output the one with the smallest V1. If there is no solution, output “No Solution” instead.

Sample Input 1:
8 15
1 2 8 7 2 4 11 15
Sample Output 1:
4 11
Sample Input 2:
7 14
1 8 7 2 4 11 15
Sample Output 2:
No Solution

解题思路：真的是PAT虐我千百遍，我待PAT如初恋，很简单的一道题，开一个面值的数组，已面值为数组下标，存放该面值的硬币的数量，然后从i=1开始遍历，看m-i的面值的硬币是否存在，如果存在，直接输出就好了，需要考虑输出两个面值一样的硬币的情况。
！！！重点来了！！！
题目中说

The second line contains N face values of the coins, which are all positive numbers no more than 500.

每个硬币的面额不超过500，那我就开了一个505的数组，我想这应该够了吧，结果测试点3死活过不了，数组开城1001才过。PAT你也真是够了，能老老实实按照题目意思来吗。

#include<iostream>
#include<stdio.h>
using namespace std;
int main(){
    for (int n, m; scanf("%d%d", &n, &m) != EOF;){
        int coins[1001] = { 0 };
        for (int i = 0; i < n; i++){
            int temp;
            scanf("%d", &temp);
            coins[temp]++;
        }
        int flag = 0;
        for (int i = 1; i < m; i++){
            if (coins[i]>0){
                coins[i]--;
                if (coins[m - i]>0){
                    flag = 1;
                    printf("%d %d\n", i, m-i);
                    break;
                }
                coins[i]++;
            }
        }
        if (!flag){
            printf("No Solution\n");
        }
    }
}