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luoguP2265 路边的水沟

题目:http://www.luogu.org/problem/show?pid=2265

题解:ans=C(n+m,n)%p

        求一下逆元就行

代码;

 1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<cmath> 5 #include<iostream> 6 #include<string> 7 #include<set> 8 #include<map> 9 #include<vector>10 #include<algorithm>11 #include<queue>12 #define for0(i,n) for(int i=0;i<=n;i++)13 #define for1(i,n) for(int i=1;i<=n;i++)14 #define for2(i,x,y) for(int i=(x);i<=(y);i++)15 #define for3(i,y,x) for(int i=(y);i>=(x);i--)16 #define maxn 100000+517 #define mod 100000000718 #define ll long long 19 using namespace std;20 inline int read()21 {22     int x=0,f=1;char ch=getchar();23     while(ch<0||ch>9){if(ch==-)f=-1;ch=getchar();}24     while(ch>=0&&ch<=9){x=10*x+ch-0;ch=getchar();}25     return x*f;26 }27 ll power(ll x,ll y)28 {29     ll t=1;30     for(;y;y>>=1,x=(x*x)%mod)31      if(y&1)t=(t*x)%mod;32     return t;33 }34 int main()35 {36     freopen("input.txt","r",stdin);37     freopen("output.txt","w",stdout);38     ll n=read(),m=read();39     ll t1=1,t2=1;40     for1(i,n)t1=(t1*(ll)i)%mod;41     for2(i,m+1,n+m)t2=(t2*(ll)i)%mod;42     printf("%lld\n",t2*power(t1,mod-2)%mod);43     return 0;44 }
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luoguP2265 路边的水沟