Vasily the Bear and Beautiful Strings

事實上這題有規律

0=1______
1=01_____
0=001____
1=0001___

可是後面底線必須一定要有數字，所以如果把原本數字的個數刪掉上面那幾個算式的個數

無法拿出任何一個數就不能再做了

當1的個數只有一個的時候，會造成上面講的問題，因為
01
001
0001
00001

所以要額外判斷這種情況

而當1的個數是零的話，也要另外判斷

//====================================================================||
//                                                                    ||
//                                                                    ||
//                         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);
    }
















}