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sequence2(高精度dp)

2015-11-22 10:51 567 查看

sequence2

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

[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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)

[align=left]Output[/align]
For each cases, please output an integer in a line as the answer.

[align=left]Sample Input[/align]

3 2
1 2 2
3 2
1 2 3

[align=left]Sample Output[/align]

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;
}


  
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