bzoj1856 [SCOI2010]字符串

题目链接

题解：http://www.cnblogs.com/jianglangcaijin/p/3443689.html

求逆元还是快速幂好用

 1 #include<algorithm>
 2 #include<iostream>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<string>
 7 #include<cmath>
 8 #include<ctime>
 9 #include<queue>
10 #include<stack>
11 #include<map>
12 #include<set>
13 #define rre(i,r,l) for(int i=(r);i>=(l);i--)
14 #define re(i,l,r) for(int i=(l);i<=(r);i++)
15 #define Clear(a,b) memset(a,b,sizeof(a))
16 #define inout(x) printf("%d",(x))
17 #define douin(x) scanf("%lf",&x)
18 #define strin(x) scanf("%s",(x))
19 #define LLin(x) scanf("%lld",&x)
20 #define op operator
21 #define CSC main
22 typedef unsigned long long ULL;
23 typedef const int cint;
24 typedef long long LL;
25 using namespace std;
26 void inin(int &ret)
27 {
28     ret=0;int f=0;char ch=getchar();
29     while(ch<'0'||ch>'9'){if(ch=='-')f=1;ch=getchar();}
30     while(ch>='0'&&ch<='9')ret*=10,ret+=ch-'0',ch=getchar();
31     ret=f?-ret:ret;
32 }
33 int n,m;
34 const LL mod=20100403;
35 LL inv(LL a,LL n)
36 {
37     a%=n;
38     LL b=n-2,ret=1;
39     while(b)
40     {
41         if(b&1)ret=ret*a%n;
42         b>>=1;
43         a=a*a%n;
44     }
45     return ret;
46 }
47 LL C(int a,int b)
48 {
49     LL x=a;
50     re(i,2,b)x=x*(a-i+1)%mod*inv(i,mod)%mod;
51     return x;
52 }
53 int CSC()
54 {
55     inin(n),inin(m);
56     printf("%lld",((C(n+m,n)-C(n+m,m-1))%mod+mod)%mod);
57     return 0;
58 }