ZOJ

题目：4个点A(0,0),B(p,0),C(m,q),D(m,n),保证p<m,q<n，求从A走到D和从B走到C两条路径不相交的走法的种数。

思路：用总的路径数-相交的路径数

相交的路径数就是A到C的路径数*B到D的路径数，因为他们之间必有交点E,A->E->C,B->E->D

代码：

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<iostream>#include<algorithm>#include<ctime>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<vector>#include<map>#include<set>#include<queue>#include<stack>#include<list>#include<numeric>using namespace std;#define LL long long#define ULL unsigned long long#define INF 0x3f3f3f3f3f3f3f3f#define mm(a,b) memset(a,b,sizeof(a))#define PP puts("*********************");template<class T> T f_abs(T a){ return a > 0 ? a : -a; }template<class T> T gcd(T a, T b){ return b ? gcd(b, a%b) : a; }template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}// 0x3f3f3f3f3f3f3f3fconst LL MOD=100000007;const int maxn=3e5+50;LL fac[maxn],Inv[maxn];LL pow_mod(LL a,LL b){    LL res=1;    while(b){        if(b&1) res=res*a%MOD;        a=a*a%MOD;        b/=2;    }    return res;}LL C(int n,int m){    return fac[n]*Inv[n-m]%MOD*Inv[m]%MOD;}int main(){    int m,n,p,q;    fac[0]=Inv[0]=1;    for(int i=1;i<maxn;i++){        fac[i]=fac[i-1]*(LL)i%MOD;        Inv[i]=pow_mod(fac[i],(LL)(MOD-2));    }    while(~scanf("%d%d%d%d",&m,&n,&p,&q)){        LL ans1=C(m+n,m)*C(m+q-p,q)%MOD;        LL ans2=C(m+q,m)*C(m+n-p,n)%MOD;        ans1=(ans1-ans2)%MOD;        ans1=(ans1+MOD)%MOD;        printf("%lld\n",ans1);    }    return 0;}