BC63 sequence2

问题描述

给定长度为$n$的序列bib_ib​i​​，求有多少长度为$k$的本质不同的上升子序列。设该序列位置为a1,a2...aka_1, a_2 ... a_k​​一个序列为上升子序列，当且仅当a1<a2<...<aka_1 < a_2 < ... < a_k​​且ba1<ba2<...<bakb_{a_1} < b_{a_2} < ... < b{a_k}。本质不同当且仅当两个序列aa和AA存在一个$i$使得ai≠Ai​​。

输入描述

若干组数据（大概$5$组）。每组数据第一行两个整数n(1≤n≤100),k(1≤k≤n)n(1 \leq n \leq 100), k(1 \leq k \leq n)。接下来一行$n$个整数bi(0≤bi≤109)b_i(0 \leq b_i \leq 10^{9})。

输出描述

对于每组的每个询问，输出一行。

输入样例

3 21 2 23 21 2 3

输出样例

23

我们可以通过$]$表示第ii个数，当前这个数为序列中的第jj个数的方案总数。转移为dp[i][j]=sumdp[k][j−1](k<i,bk<bi)dp[i][j] = sum{dp[k][j-1]}(k < i, b_k < b_i)。本题需要高精度。

这道题只需要用到大数加法就行了，

#include <stdio.h>#include <string>#include <iostream>using namespace std;const int maxn = 105;string dp[maxn][maxn], ch[maxn];//一维表示j-i的长度序列,二维表示长度int a[maxn];void swap ( string & ch ){    for ( int i = 0, j = ch.size ( )-1; i < j; i ++, j -- )    {        char t = ch[i];        ch[i] = ch[j];        ch[j] = t;    }}void Add ( string & str, const string ch )  //大数加法{    string s = "";    int i, j, carry = 0, t;    for ( i = str.size ( )-1, j = ch.size ( )-1; i >= 0 && j >= 0; i --, j -- )    {        t = ( str[i]-'0' )+( ch[j]-'0' )+carry;        carry = t/10;        t = t%10;        s = s+( char )( t+'0' );    }    while ( i >= 0 )    {        t = ( str[i]-'0' )+carry;        carry = t/10;        t = t%10;        s = s+( char )( t+'0' );        i --;    }    while ( j >= 0 )    {        t = ( ch[j]-'0' )+carry;        carry = t/10;        t = t%10;        s = s+( char )( t+'0' );        j --;    }    if ( carry > 0 )    //进位        s = s+( char )( carry+'0' );    swap ( s );    str = s;}int main ( ){    int n, k;    string ans;    while ( ~ scanf ( "%d%d", &n, &k ) )    {        for ( int i = 1; i <= n; i ++ )            scanf ( "%d", &a[i] );        for ( int i = 1; i < maxn; i ++ )        {            for ( int j = 1; j < maxn; j ++ )                dp[i][j] = "0";            dp[i][1] = "1";        }        for ( int i = 1; i <= n; i ++ )        {            for ( int j = 1; j < i; j ++ )            {                if ( a[j] >= a[i] ) //找到前面比a[i]小的a[j]                    continue ;                for ( int l = 2; l <= k; l ++ )                //将递增序列长度为2-k全部累加                    Add ( dp[i][l], dp[j][l-1] );            }        }        ans = "0";        for ( int i = 1; i <= n; i ++ )            Add ( ans, dp[i][k] );  //加长度为的k的值加起来        cout << ans << endl;    }    return 0;}