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  • HDU 6168 Numbers(模拟)

    Numbers

    HDU 6168 Numbers 2017ACM暑期多校联合训练 - Team 9 1008

    题目链接

    Problem Description
    zk has n numbers a1,a2,…,an. For each (i,j) satisfying 1≤i<j≤n, zk generates a new number (ai+aj). These new numbers could make up a new sequence b1,b2,…,bn(n−1)/2.
    LsF wants to make some trouble. While zk is sleeping, Lsf mixed up sequence a and b with random order so that zk can’t figure out which numbers were in a or b. “I’m angry!”, says zk.
    Can you help zk find out which n numbers were originally in a?

    Input
    Multiple test cases(not exceed 10).
    For each test case:
    ∙The first line is an integer m(0≤m≤125250), indicating the total length of a and b. It’s guaranteed m can be formed as n(n+1)/2.
    ∙The second line contains m numbers, indicating the mixed sequence of a and b.
    Each ai is in [1,10^9]

    Output
    For each test case, output two lines.
    The first line is an integer n, indicating the length of sequence a;
    The second line should contain n space-seprated integers a1,a2,…,an(a1≤a2≤…≤an). These are numbers in sequence a.
    It’s guaranteed that there is only one solution for each case.

    Sample Input
    6
    2 2 2 4 4 4
    21
    1 2 3 3 4 4 5 5 5 6 6 6 7 7 7 8 8 9 9 10 11

    Sample Output
    3
    2 2 2
    6
    1 2 3 4 5 6

    题意:

    已知数组a由n个元素组成,数组b由数组a里面的任意两个元素相加组成,则数组b里面的元素个数为n(n-1)/2.

    现在a数组和b数组混合在一起一共有m个元素,求解a数组里面的元素个数,并将这些元素打印出来。

    模拟,混合数组里面的两个最小的元素一定是a数组里面的值,这样每次将找到的值与a数组中已有的值相加得到的值从混合数组里面删去,这样混合数组中现存的第一个值肯定就是要加入到a数组里面的,接着重复进行上面的过程,直到找齐a数组里面的元素。

    #include <bits/stdc++.h>
    using namespace std;
    const int maxn = 1e4;
    typedef long long ll;
    
    ll a[maxn];
    int main()
    {
        ll num, m;
        while(~scanf("%lld", &m))
        {
            map <ll, ll> as;
            ll min1=1000000009,min2=1000000000;
            ll n=(sqrt(8*m+1)-1)/2;
            for(ll i=0; i<m; i++)
            {
                scanf("%lld",&num);
                if(num<min2)
                {
                    min1=min2;
                    min2=num;
                }
                else if(num<min1)
                {
                    min1=num;
                }
                as[num]++;
            }
            as[min1] --;
            as[min2] --;
            a[0] = min2;
            a[1] = min1;
            ll ans = 1;
            while(ans != n - 1)
            {
                for(ll i = 0; i < ans; ++ i)///前ans个元素于第ans个元素想加,再在map中减去
                {
                    as[a[i] + a[ans]] --;
                }
                map<ll, ll>::iterator it = as.begin();
                while(it!= as.end())
                {
                    if(it -> second == ll(0))///如果该元素个数为零,将该元素删除
                    {
                        as.erase(it ++);
                    }
                    else
                    {
                        break;
                    }
                }
                a[++ ans] = as.begin() -> first;///a数组元素加一
                as[a[ans]] --;
            }
            printf("%lld
    ",n);
            printf("%lld",a[0]);
            for(int i=1; i<n; i++)
                printf("% lld",a[i]);
            printf("
    ");
        }
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/nanfenggu/p/7900040.html
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