1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>

using namespace std;

typedef vector<int> VI;
typedef long long LL;

#define FOR(x, b, e) for(int x = b; x <= (e); ++x)
#define FORD(x, b, e) for(int x = b; x >= (e); --x)
#define REP(x, n) for(int x = 0; x < (n); ++x)
#define VAR(v, n) __typeof(n) v = (n)
#define ALL(c) (c).begin(), (c).end()
#define SIZE(x) ((int)(x).size())
#define FOREACH(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i)
#define PB push_back
#define ST first
#define ND second


string dod(string a, string b){
    string res=a;
    int temp = 0, p = 0;
    FORD(i,SIZE(a)-1,0){
        temp = a[i]+b[i]+p-'0';
        p = 0;
        if(temp>57) {
            temp-=10;
            p+=1;
        }
        res[i] = (char)temp;

    }


    return res;
}

int main()
{
    string a, b, c, res, pom1, pom2, pom3;
    cin>>a>>b>>c;
//    res = dod(a,b);
    int k = 0, odp = 0;
    FOR(k,1, SIZE(a)){
        REP(i, SIZE(a)-k+1){
            pom1 ='0'+ a.substr(i,k);
            pom2 ='0'+ b.substr(i,k);
            pom3 ='0'+ c.substr(i,k);
           // cout<<pom1<<" + "<<pom2<<" ?= "<<pom3<<endl;
            res = dod(pom1,pom2);
            if(res == pom3) {odp++;
           //     cout<<pom1<<" + "<<pom2<<" = "<<pom3<<endl;
            }
        }
    }
    cout<<odp<<endl;

    return 0;
}