sequence2(高精度dp)

sequence2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 220    Accepted Submission(s): 90

Problem Description

Given an integer array bi with a length of n, please tell me how many exactly different increasing subsequences.
P.S. A subsequence bai(1≤i≤k) is an increasing subsequence of sequence bi(1≤i≤n) if and only if 1≤a1<a2<...<ak≤n and ba1<ba2<...<bak. Two sequences ai and bi is exactly different if and only if there exist at least one i and ai≠bi.

 

Input

Several test cases(about 5)
For each cases, first come 2 integers, n,k(1≤n≤100,1≤k≤n)
Then follows n integers ai(0≤ai≤109)

 

Output

For each cases, please output an integer in a line as the answer.

 

Sample Input

3 2 1 2 2 3 2 1 2 3

 

Sample Output

2 3

题解：让求一个数列中长度为k的LIS数列的种数（指的数组下标）；所以想到用dp，二维dp，dp[i][j]其中i指的是长度，j指的是以j结束的数；所以可以列出状态转移方程；

dp[x][i]=dp[x-1][j]+dp[x][i];每当if(m[i]>m[j])时 开始从2到n遍历；由于数量太大，所以要用到高精度。。。

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
typedef long long LL;
struct BIGINT{
    int num[20],len;
    void init(int x){
        mem(this->num,0);
        this->num[0]=x;
        this->len=1;
    }
};
BIGINT operator + (BIGINT a,BIGINT b){
    BIGINT c;
    c.init(0);//c要记得初始化。。。 
    int len=max(a.len,b.len);
    for(int i=0;i<len;i++){
        c.num[i]=a.num[i]+b.num[i]+c.num[i];
        if(c.num[i]>1e8)c.num[i]-=1e8,c.num[i+1]++;
        if(c.num[len])len++;
    }c.len=len;
    return c;
}
void print(BIGINT a){
    for(int i=a.len-1;i>=0;i--){
        printf("%d",a.num[i]);
    }puts("");
}
BIGINT dp[110][110],ans;//以j结尾长度为i的个数 
int m[110];
int main(){
    int n,k;
    while(~scanf("%d%d",&n,&k)){
        for(int i=1;i<=n;i++)scanf("%d",m+i);
        mem(dp,0);
        for(int i=1;i<=n;i++)dp[1][i].init(1);
        for(int i=1;i<=n;i++)
        for(int j=1;j<i;j++)
        if(m[i]>m[j]){
            for(int x=2;x<=n;x++){
                dp[x][i]=dp[x-1][j]+dp[x][i];
            }
        }
        ans.init(0);
        for(int i=1;i<=n;i++)ans=ans+dp[k][i];
        print(ans);
    }
    return 0;
}

大神优化过的代码。。。好难懂。。。还没懂。。。

代码：

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define ull unsigned long long
#define ll long long
#define N 100005
#define BASE 13131
int n,k;
int a[105];
struct Cbig
{
    int num[20],len;
    void init(int x)
    {
        len=1;
        num[0]=x;
    }
}dp[2][105],ans,c;
Cbig add(Cbig &a,Cbig &b)
{
    c.len=max(a.len,b.len);
    c.num[0]=0;
    for(int i=0;i<c.len;i++)
    {
        c.num[i+1]=0;
        if(i<a.len) c.num[i]+=a.num[i];
        if(i<b.len) c.num[i]+=b.num[i];
        if(c.num[i]>=100000000)
        {
            c.num[i]-=100000000;
            c.num[i+1]=1;
        }
    }
    if(c.num[c.len]) c.len++;
    return c;
}
void print(Cbig &a)
{
    printf("%d",c.num[a.len-1]);
    for(int i=a.len-2;i>=0;i--)
        printf("%08d",a.num[i]);
    puts("");
}
int main()
{
    //freopen("tt.in", "r", stdin);
    while(cin>>n>>k)
    {
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        dp[0][0].init(1);
        for(int i=1;i<=n;i++) dp[0][i].init(0);
        int p=0,q=1;
        for(int t=1;t<=k;t++)
        {
            p^=1;q^=1;
            for(int i=0;i<=n;i++) dp[p][i].init(0);
            for(int i=1;i<=n;i++)
            {
                for(int j=0;j<i;j++)
                if(j==0||a[j]<a[i])
                    dp[p][i]=add(dp[p][i],dp[q][j]);
            }
        }
        ans.init(0);
        for(int i=1;i<=n;i++)
            ans=add(ans,dp[p][i]);
        print(ans);
    }
    return 0;
}