Codeforces Round #564 (Div. 2)A_nauuo is a girl who loves writing comments. one da-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_43276292/article/details/91349539

本文解析了Codeforces上一道竞赛题目，关于Nauuo的评论可能获得的投票结果。通过分析已知的点赞、踩踏及未知投票者的数量，确定最终的投票结果是否确定，或者存在不确定性。

A. Nauuo and Votes

题目链接：http://codeforces.com/contest/1173/problem/A

题目

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 xpersons who would upvote, yy y persons who would downvote, and there were also another z persons who would vote, but you don’t know whether they would upvote or downvote. Note that each of the x+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 z z 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 z z persons vote, that the results are different in the two situations.

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

Input

The only line contains three integers x,y,z(0=<x,y,z<=100) x, corresponding to the number of persons who would upvote, downvote or unknown.

Output

If there is only one possible result, print the result : “+”, “-” or “0”.
Otherwise, print “?” to report that the result is uncertain.
Example

intput

3 7 0

output

题意

There are there numbers , they are x , y and z. you can give z to x and y，

if x are the most ,you can output “+”，

if y are the most ,you can output “-”，

if you are not sure which are the most,you should output “?”.

思路

it’s a easy problem,you can compare a and b+c,b and a+c.

#include<bits/stdc++.h>
using namespace std;
int main()
{

    int a,b,c;
    while(cin>>a>>b>>c)
    {
        if(a==b&&c==0)
            cout<<0<<endl;
        else if(a>b+c)
            cout<<"+"<<endl;
        else if(b>a+c)
            cout<<"-"<<endl;
        else
            cout<<"?"<<endl;
    }
    return 0;
}