hdu1223

来源：互联网发布：java发展前景编辑：程序博客网时间：2024/06/06 23:56

n个数，只能用<或者=连接，共有多少种方案。

F[i][j]=(F[i-1][j]+F[i-1][j-1])*j

F[i][j]代表i个数，有j个不同值的情况。比如A<B=C<D,j=3.

大数模板

#include <stdio.h>#include <string.h>const int MAX =505;struct BigNum{    int num[MAX];    int len;} a[51][51];BigNum  Add(BigNum &a, BigNum &b){    BigNum c;    int  i, len;    len = (a.len > b.len) ? a.len : b.len;    memset(c.num, 0, sizeof(c.num));    for(i = 0; i < len; i++)    {        c.num[i] += (a.num[i]+b.num[i]);        if(c.num[i] >= 10)        {            c.num[i+1]++;            c.num[i] -= 10;        }    }    if(c.num[len]) len++;    c.len = len;    return  c;}BigNum Mul1(BigNum &a, int  &b){    BigNum c;    int  i, len;    len = a.len;    memset(c.num, 0, sizeof(c.num));    //乘以0，直接返回0    if(b == 0)    {        c.len = 1;        return  c;    }    for(i = 0; i < len; i++)    {        c.num[i] += (a.num[i]*b);        if(c.num[i] >= 10)        {            c.num[i+1] = c.num[i]/10;            c.num[i] %= 10;        }    }    while(c.num[len] > 0)    {        c.num[len+1] = c.num[len]/10;        c.num[len++] %= 10;    }    c.len = len;    return  c;}void init(){    int i,j;    for(i=0; i<=50; i++)        for(j=0; j<=50; j++)            if(i<j||j==0)            {                a[i][j].num[0]=0;                a[i][j].len=1;            }            else if(j==1)            {                a[i][j].num[0]=1;                a[i][j].len=1;            }            else            {                a[i][j]=Add(a[i-1][j],a[i-1][j-1]);                a[i][j]=Mul1(a[i][j],j);            }}int main(){    BigNum sum;    int i,p,n;    init();    scanf("%d",&p);    while(p--)    {        for(i=0;i<MAX;i++)            sum.num[i]=0;        sum.len=0;        scanf("%d",&n);        for(i=1;i<=n;i++)            sum=Add(sum,a[n][i]);        for(i=sum.len-1;i>=0;i--)            printf("%d",sum.num[i]);        printf("\n");    }    return 0;}

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