Codeforces Round #564 (Div. 2) A. Nauuo and Votes

Codeforces竞赛A题解析
本文解析了Codeforces竞赛中的一道题目,讨论了一个女孩在平台上发表评论后可能收到的点赞和踩踏情况,通过数学分析确定结果是否确定,提供了一段C++代码实现。

链接:https://codeforces.com/contest/1173/problem/A

题意:

Nauuo is a girl who loves writing comments.

One day, she posted a comment on Codeforces, wondering whether she would get upvotes or downvotes.

It's known that there were xx persons who would upvote, yy persons who would downvote, and there were also another zz persons who would vote, but you don't know whether they would upvote or downvote. Note that each of the x+y+zx+y+z people would vote exactly one time.

There are three different results: if there are more people upvote than downvote, the result will be "+"; if there are more people downvote than upvote, the result will be "-"; otherwise the result will be "0".

Because of the zz unknown persons, the result may be uncertain (i.e. there are more than one possible results). More formally, the result is uncertain if and only if there exist two different situations of how the zz persons vote, that the results are different in the two situations.

Tell Nauuo the result or report that the result is uncertain.

思路:

x-y的绝对值大于z就有稳定的答案,想同时特判z等于0.

代码:

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
const int MAXN = 1e5 + 10;
const int MOD = 1e9 + 7;
int n, m, k, t;

int main()
{
//    freopen("test.in", "r", stdin);
    int x, y, z;
    cin >> x >> y >> z;
    if (z < abs(x - y) || z == 0)
    {
        if (x > y)
            cout << "+" << endl;
        else if (y > x)
            cout << "-" << endl;
        else
            cout << "0" << endl;
    }
    else
        cout << "?" << endl;

    return 0;
}

  

转载于:https://www.cnblogs.com/YDDDD/p/10990861.html

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