zoukankan      html  css  js  c++  java
  • 1048. Find Coins (25)

    1048. Find Coins (25)

    时间限制
    100 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
    


    代码:

    #include <iostream>
    #include <algorithm>
    using namespace std;
    
    int main()
    {
        int n,m,flag=0;
        int a[100000];
        cin>>n>>m;
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
            if(a[i]>=m)i--,n--;
        }
        sort(a,a+n);
        for(int i=n-1;i>=0;i--)
        {
            if(a[i]+a[0]>m)n--;
            else break;
        }
        if(a[n-1]+a[n-2]>=m)
        for(int i=0;i<n-1;i++)
        {
            if(a[i]+a[i+1]>m)break;
            else if(a[i]+a[n-1]<m)continue;//加上这一条就不超时了。。
            for(int j=i+1;j<n;j++)
            if(a[i]+a[j]==m)
            {
                cout<<a[i]<<' '<<a[j];
                flag=1;
                break;
            }
            if(flag)break;
        }
        if(flag==0)cout<<"No Solution";
    }
  • 相关阅读:
    [转]优秀的程序员不会觉得累成狗是一种荣耀
    .NET读写Excel工具Spire.XlS使用(DataExport )
    WPF之Binding深入探讨
    第一个WPF应用程序
    Visio作图
    唯一的重复元素
    Strange Problem O(∩_∩)O~
    数据库知识点①
    HDU 2825 Wireless Password
    POJ 1625 Censored!
  • 原文地址:https://www.cnblogs.com/8023spz/p/7299468.html
Copyright © 2011-2022 走看看