Nauuo and Votes

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

题解：思维

1、当且仅当x>y并且x>y+z时，upvote；

2、当且仅当x==y并且z==0时，0；

3、当且仅当x<y并且x+z<y时，downvote；

4、其余情况都是？；

5、由于z如果全部加到劣势的一方，依然不能赢的话，优势一方必赢；

/*
*@Author:   STZG
*@Language: C++
*/
#include <bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<vector>
#include<bitset>
#include<queue>
#include<deque>
#include<stack>
#include<cmath>
#include<list>
#include<map>
#include<set>
//#define DEBUG
#define RI register int
#define endl "\n"
using namespace std;
typedef long long ll;
//typedef __int128 lll;
const int N=100000+10;
const int M=100000+10;
const int MOD=1e9+7;
const double PI = acos(-1.0);
const double EXP = 1E-8;
const int INF = 0x3f3f3f3f;
int t,n,m,k,p,l,r,u,v;
int ans,cnt,flag,temp,sum;
int a[N];
char str;
struct node{};
int main()
{
#ifdef DEBUG
	freopen("input.in", "r", stdin);
	//freopen("output.out", "w", stdout);
#endif
    //ios::sync_with_stdio(false);
    //cin.tie(0);
    //cout.tie(0);
    //scanf("%d",&t);
    //while(t--){
    scanf("%d%d%d",&n,&m,&k);
    u=(n>m)&&(n>m+k);
    p=(n==m)&&(k==0);
    v=(n<m)&&(n+k<m);
    cout<<(p?'0':(u||v?(u?'+':'-'):'?'))<<endl;
    //}

#ifdef DEBUG
	printf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC);
#endif
    //cout << "Hello world!" << endl;
    return 0;
}