问题描述

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 V₁and 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.

解决方法

#include<cstdio>
#include<algorithm>
using namespace std;
const int maxn = 100005;
int arr[maxn];
int main(void)
{
  int m = 0,n = 0;
  bool flag = false;
  scanf("%d %d",&m,&n);
  for(int i = 0;i<m;i++)
  {
    scanf("%d",arr+i);
  }
  sort(arr,arr+m);
  //使用两个指针指向数组开头和结尾处向内前进
  int pre = 0,suf = m - 1;
  while(pre!=suf)
  {
    if(arr[pre] + arr[suf] > n)
    {
      suf--;
    }else if(arr[pre]+arr[suf] < n)
    {
      pre++;
    }else
    {
      flag = true;
      break;
    }
  }
  if(flag)
  {
    printf("%d %d",arr[pre],arr[suf]);
  }else
  {
    printf("No Solution");
  }
  return 0;
}

基本策略

• 使用两个指针指向数组开始和结束通过比较两者之和和要求数的结果来决定那个指针移动，如果相等就直接跳出输出结果即可。如果到结束都没发现则输出No Solution

结果展示

PatA1048.png