September 10th 模拟赛C T1 电影票 Solution

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Description

笨笨当了很久的道路调度员，笨笨也开始想体验生活，从生活中发现数学问题，锻炼自己思维。最近《变形金刚3》，《哈利波特7》同步放映，明显是决战雌雄，已知王府井中一共有n人买了《变形金刚3》的票，m人买了《哈利波特7》的票，并且n>=m，并且电影院中现在只有两种票，每次只有一个人买，(共有n+m次)，这n+m次组成一个排列，为了保证每一个人买票时，《变形金刚3》票房都不少于《哈利波特7》，(n个买《变形金刚3》的人之间没区别，m个买《哈利波特7》的人也没区别),笨笨想着到这样的购票方案有多少种。笨笨想了好久都没想出来，所以笨笨找到了你。

Input

一行两个数n,m ( 0<=m<=n<=5000)

Output

输出方案种数

Solution

套上神器的公式:Ans=(N+2)×(N+3)×⋯×(N+M−1)×(N+M)×(N+1−M)M!
此处不予证明.

Code

C++

#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#include <cstdlib>#include <cmath>#define F(i,x,y) for(long long i=x;i<=y;++i)using namespace std;struct node{    int len,a[3500];    node()    {        len=1;    }};int num=0;int prime[1501],a[10010];    int isNotPrime[10010] = {1, 1};    node chengfa_x(node n1,int x){    int i;node no;    no.len=n1.len;    for(i=1;i<=no.len;i++) no.a[i]=n1.a[i]*x;    for(i=1;i<=no.len;i++)    {        no.a[i+1]+=no.a[i]/10;        no.a[i]%=10;    }    i=no.len;    while(no.a[i+1]>0)    {        i++;        no.a[i+1]+=no.a[i]/10;        no.a[i]%=10;    }    while(no.a[i]==0&&i>1) i--;    no.len=i;    return no;}void workcheng(int x){    int t=x;    F(i,0,num)    {            while(t%prime[i]==0)             {                t/=prime[i];                a[i]++;            }        if(t<=1) break;    }}void workchu(int x){    int t=x;    F(i,0,num)    {            while(t%prime[i]==0)             {                t/=prime[i];                a[i]--;            }        if(t<=1) break;    }}int main(){    int i,j,r,n,m,k,l,w;    node ans;    ans.a[1]=1;    scanf("%d%d",&n,&m);    for(long i = 2 ; i <  10000 ; i ++)        {            if(! isNotPrime[i]) prime[num++]=i;            for(long j = 0 ; j < num && i * prime[j] <  10000 ; j ++)            {                isNotPrime[i * prime[j]] = 1;                if( !(i % prime[j])) break;            }        }        F(i,2,m) workcheng(n+i);    workcheng(n+1-m);    F(i,2,m) workchu(i);    F(i,0,num)     {        F(j,1,a[i]) ans=chengfa_x(ans,prime[i]);    }    for(i=ans.len;i>=1;i--) printf("%d",ans.a[i]);    return 0;}

Pascal

type    hp=array [0..3000] of int64;var    ans:hp;    i,n,m:longint;function times(x:hp;y:longint):hp;var    i:longint;begin    fillchar(times,sizeof(times),0);    for i:=1 to x[0] do    begin        inc(times[i],x[i]*y);        times[i+1]:=times[i] div 10000000;        times[i]:=times[i] mod 10000000;    end;    times[0]:=x[0];    if times[x[0]+1]<>0 then        inc(times[0]);end;function divide(x:hp;y:longint):hp;var    i,t:longint;begin    fillchar(divide,sizeof(divide),0);    t:=0;    for i:=x[0] downto 1 do    begin        divide[i]:=(x[i]+t*10000000) div y;        t:=(x[i]+t*10000000) mod y;    end;    divide[0]:=x[0];    while divide[divide[0]]=0 do        dec(divide[0]);end;procedure print(x:hp);var    i,j:longint;begin    write(x[x[0]]);    for i:=x[0]-1 downto 1 do    begin        j:=1000000;        while (x[i]<j) and (j>1) do        begin            write(0);            j:=j div 10;        end;        write(x[i]);    end;    writeln;end;begin    readln(n,m);    ans[0]:=1;    ans[1]:=n+1-m;    for i:=n+2 to n+m do        ans:=times(ans,i);    for i:=1 to m do        ans:=divide(ans,i);    print(ans);end.