In fact, this problem has a pattern.
0=1______
1=01_____
0=001____
1=0001___
However, the underscores at the end must have a number, so if you remove the number of original digits from the number of the above equations,
If you cannot take out any number, you cannot proceed further.
When there is only one '1', it will cause the problem mentioned above, because
01
001
0001
00001
So you need to additionally judge this situation.
And when the number of '1's is zero, it also needs to be judged separately.
//====================================================================||
// ||
// ||
// Author : GCA ||
// 6AE7EE02212D47DAD26C32C0FE829006 ||
//====================================================================||
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
#include <ctime>
using namespace std;
#ifdef DEBUG
#define debug(...) printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#define gettime() end\_time=clock();printf("now running time is %.7f\\n",(float)(end\_time - start\_time)/CLOCKS\_PER\_SEC);
#else
#define debug(...)
#endif
typedef unsigned int uint;
typedef long long int Int;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\\n")
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()
#define De(x)printf(#x"%d\\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\
printf("%d ",dp\[htx\]\[hty\]);}Pln();}
#define M 100005
#define PII pair<int,int\>
#define PB push\_back
#define oo INT\_MAX
#define Set\_oo 0x3f
#define FOR(it,c) for(\_\_typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define eps 1e-6
clock\_t start\_time=clock(), end\_time;
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
int tmp=min(x,y);
return min(tmp,z);
}
int max3(int x,int y,int z){
int tmp=max(x,y);
return max(tmp,z);
}
Int n,m,g;
Int f\[M+M+10\];
Int mod=1000000007;
void pre(){
f\[0\]=1;
for(int i=1;i<M+M+5;i++){
f\[i\]=i\*f\[i-1\];
f\[i\]%=mod;
}
}
Int gpow(Int x,Int p){
Int ans=1;
while(p>0){
if(p&1)
ans=ans\*x%mod;
x=x\*x%mod;
p>>=1;
}
return ans;
}
Int inv(Int x){
return gpow(x,mod-2);
}
Int c(int n,int p){
debug("%d %d\\n",n,p);
if(n==p)return 1;
if(p==0)return 1;
Int ans=f\[n\]\*inv(f\[p\]\*f\[n-p\]%mod);
ans%=mod;
debug("aa %d\\n",ans);
return ans;
}
int main() {
ios\_base::sync\_with\_stdio(0);
pre();
while(~scanf("%I64d%I64d%I64d",&n,&m,&g)){
Int ans=0;
if(g){
if(m){
for(int i=1;i<=n;i+=2){
if(n+m-i-1<=0)break;
ans+=c(n+m-i-1,n-i);
ans%=mod;
}
if(m==1)ans=(ans+((!(n&1))?1:0))%mod;
}else{
ans=(!(n&1))?1:0;
}
}else{
if(m){
for(int i=0;i<=n;i+=2){
if(n+m-i-1<=0)break;
ans+=c(n+m-i-1,n-i);
ans%=mod;
}//debug("2a%d\\n",ans);
if(m==1)ans=(ans+((n&1)?1:0))%mod;
}else{
ans=((n&1))?1:0;
}
}
printf("%I64d\\n",ans);
}
}