More than 2019 cattle off summer school camp (first) ABBA

That Italy: (n + m) th A, (n + m) th B composed of a string, which was cut open, so that a number of n and m AB BA solution

Thinking 1: DP, dp [i] [j] denotes the i-th beginning with A, j at the beginning of the program number of B.

　　     If this current is A, then it should satisfy, ij <= n, the number of AB not exceed (n + 1)

           Similarly in the case of B.

 

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=b;i>=a;i--)
using namespace std;
#define ll long long
const int N=3e5+5;
const int mod = 1e9+7;
int n,m,f[3010][3010];
ll rd()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int sum(int a, int b) {
    int s = (a + b);
    if (s >= mod) s -= mod;
    return s;
}
void init()
{

    rep(i,0,n+m)
        rep(j,0,n+m) f[i][j]=0;
    f[0][0]=1;
    rep(i,0,n+m)
        rep(j,0,n+m)
        {
            if(i-j<=n&&i!=0) f[i][j]=sum(f[i][j],f[i-1][j]);
            if(j-i<=m&&j!=0) f[i][j]=sum(f[i][j],f[i][j-1]);
            //printf("i=%d j=%d %d\n",i,j,f[i][j]);
        }
    printf("%d\n",f[n+m][n+m]);
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        init();    
    } 
} 

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Ideas 2: the number of combinations

　　　 years = C (2 * (n + m), n + m) - C (2 * (n + m), n-1) -C (2 * (n + m), m-1)

　　　 C (2 * (n + m), n-1) is the meaning

            We have (n + m + m + 1) th A, and (n-1) th B, rearranges what program number.

　　　 Select difference number A and B is (2 * m + 2) prefix, AB interchange, is not legal answers.

#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define dep(i,a,b) for(int i=b;i>=a;i--)
using namespace std;
#define ll long long
const int N=3e5+5;
const int mod = 1e9+7;
int n,m;
ll fac[10100],inv[10100];
int sum(int a, int b) {
    int s = (a + b);
    if (s >= mod) s -= mod;
    return s;
}
int sub(int a, int b) {
    int s = a - b;
    if (s < 0) s += mod;
    return s;
}
int mult(int a, int b) {
    return (1LL * a * b) % mod;
}
ll rd()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
ll qpow(ll a,ll b,ll p)
{
    ll ret=1;a%=p;
    while(b)
    {
        if(b&1) ret=ret*a%p;
        b/=2;a=a*a%p;
    }
    return ret;
}
ll C(ll n,ll m)
{
    if(m>n) return 0;
    return 1ll * fac[n] * inv[m] % mod * inv[n - m] % mod;
}
int main()
{
    fac[0]=1;inv[0]=1;
    for(int i=1;i<=10000;i++) 
    {
        fac[i]=1ll*fac[i-1]*i%mod;
        inv[i]=qpow(fac[i],mod-2,mod);    
    }
    while(~scanf("%d%d",&n,&m))
    {
        printf("%d\n",sub(sub(C(2*(n+m),n+m),C(2*(n+m),n-1)),C(2*(n+m),m-1)));
    }
} 

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