1048. Find Coins (25)

最新推荐文章于 2021-02-09 21:24:45 发布

原创最新推荐文章于 2021-02-09 21:24:45 发布 · 392 阅读

0 ·

CC 4.0 BY-SA版权

PATA 专栏收录该内容

55 篇文章

订阅专栏

本博客介绍了一个算法问题，即如何使用不超过105个面额不超过500的硬币，通过找到任意金额的两个硬币组合来支付指定金额。详细解释了排序和双指针的方法来解决此问题。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

1048. Find Coins (25)

时间限制

50 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

题意:给定一串数,问能否在其中找到两个数,恰好等于m

题解:多种做法,我提供一种

首先排序,然后两个指针分别从两边开始,

如果两个数相加结果等于m,找到了,退出

如果和大于m,则右边的指针左移

如果和小于m,则左边的指针右移

#include <cstdio>
#include <algorithm>
using namespace std;

int main() {
    int ary[100010];
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = 0; i < n; ++i) 
        scanf("%d", ary + i);
    sort(ary, ary + n);
    int i = 0, j = n - 1;
    while (i < j) {
        if (ary[i] + ary[j] == m) {
            printf("%d %d\n", ary[i], ary[j]);
            break;
        }
        else if (ary[i] + ary[j] > m)
            --j;
        else 
            ++i;
    }
    if (i == j) 
        puts("No Solution");
    return 0;
}