BZOJ2729 [HNOI2012]排队（排列组合+高精度）

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【题解】

A(n,n)*( A(2,n+1)*A(m,n+3) + C(1,n+1)*A(2,2)*C(1,m)*A(m-1,n+2) )

= A(n,n)*( A(2,n+1)*A(m,n+3) + (n+1)*2*m*A(m-1,n+2) )

为防止数字过长导致复杂度太大，用long long压9位(防止乘爆)

【代码】

#include<stdio.h>#include<stdlib.h>#include<string.h>#define LEN 2000#define MOD 1000000000llstruct big{long long s[LEN+5];big(){memset(s,0,sizeof(s));}};int get(big a){int i;for(i=1;i<LEN;i++)if(a.s[i]!=0) break;return i;}void jw(big& a){int i;for(i=LEN;i>0;i--){a.s[i-1]+=a.s[i]/MOD;a.s[i]%=MOD;}}big cheng_num(big a,int b){int i;for(i=LEN;i>0;i--)a.s[i]*=b;jw(a);return a;}big cheng(big a,big b){big res;int i,j,ta=get(a),tb=get(b);for(i=LEN;i>=tb;i--){for(j=LEN;j>=ta;j--)res.s[i+j-LEN]+=b.s[i]*a.s[j];jw(res);}return res;}big jia(big a,big b){big res;int i;for(i=LEN;i>0;i--)res.s[i]=a.s[i]+b.s[i];jw(res);return res;}big A(int m,int n){big res;int i;res.s[LEN]=1;for(i=n-m+1;i<=n;i++)res=cheng_num(res,i);return res;}void print(big a){int i,ta=get(a);printf("%lld",a.s[ta]);for(i=ta+1;i<=LEN;i++)printf("%09lld",a.s[i]);}int main(){big t;int n,m,i,j,k;scanf("%d%d",&n,&m);if(m>n+3) printf("0");else print(   cheng(  A(n,n)  ,  jia( cheng(A(2,n+1),A(m,n+3)) , cheng_num(A(m-1,n+2),(n+1)*2*m) )  )   );return 0;}

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