BestCoder Round #63 (div.2) HDOJ5568 sequence2(dp + 大数)

sequence2

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 231    Accepted Submission(s): 95


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

 

题目链接:点击打开链接

动态规划题目, dp[i][j]表示序列长度为i序列最后一个数下标为j的个数, 注意要使用大数, 状态转移方程为: dp[i][j] = dp[i][j] + dp[i - 1][k]

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
#include "list"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int MAXN = 9999;
const int MAXSIZE = 10;
const int DLEN = 4;
class BigNum
{
private:
int a[500]; //可以控制大数的位数
int len; //大数长度
public:
BigNum(){ len = 1;memset(a,0,sizeof(a)); } //构造函数
BigNum(const int); //将一个int类型的变量转化为大数
BigNum(const char*); //将一个字符串类型的变量转化为大数
BigNum(const BigNum &); //拷贝构造函数
BigNum &operator=(const BigNum &); //重载赋值运算符，大数之间进行赋值运算

friend istream& operator>>(istream&, BigNum&); //重载输入运算符
friend ostream& operator<<(ostream&, BigNum&); //重载输出运算符

BigNum operator+(const BigNum &) const; //重载加法运算符，两个大数之间的相加运算
BigNum operator-(const BigNum &) const; //重载减法运算符，两个大数之间的相减运算
BigNum operator*(const BigNum &) const; //重载乘法运算符，两个大数之间的相乘运算
BigNum operator/(const int &) const; //重载除法运算符，大数对一个整数进行相除运算

BigNum operator^(const int &) const; //大数的n次方运算
int operator%(const int &) const; //大数对一个int类型的变量进行取模运算
bool operator>(const BigNum & T)const; //大数和另一个大数的大小比较
bool operator>(const int & t)const; //大数和一个int类型的变量的大小比较

void print(); //输出大数
};
BigNum::BigNum(const int b) //将一个int类型的变量转化为大数
{
int c,d = b;
len = 0;
memset(a,0,sizeof(a));
while(d > MAXN)
{
c = d - (d / (MAXN + 1)) * (MAXN + 1);
d = d / (MAXN + 1);
a[len++] = c;
}
a[len++] = d;
}
BigNum::BigNum(const char*s) //将一个字符串类型的变量转化为大数
{
int t,k,index,l,i;
memset(a,0,sizeof(a));
l=strlen(s);
len=l/DLEN;
if(l%DLEN)
len++;
index=0;
for(i=l-1;i>=0;i-=DLEN)
{
t=0;
k=i-DLEN+1;
if(k<0)
k=0;
for(int j=k;j<=i;j++)
t=t*10+s[j]-'0';
a[index++]=t;
}
}
BigNum::BigNum(const BigNum & T) : len(T.len) //拷贝构造函数
{
int i;
memset(a,0,sizeof(a));
for(i = 0 ; i < len ; i++)
a[i] = T.a[i];
}
BigNum & BigNum::operator=(const BigNum & n) //重载赋值运算符，大数之间进行赋值运算
{
int i;
len = n.len;
memset(a,0,sizeof(a));
for(i = 0 ; i < len ; i++)
a[i] = n.a[i];
return *this;
}
istream& operator>>(istream & in, BigNum & b) //重载输入运算符
{
char ch[MAXSIZE*4];
int i = -1;
in>>ch;
int l=strlen(ch);
int count=0,sum=0;
for(i=l-1;i>=0;)
{
sum = 0;
int t=1;
for(int j=0;j<4&&i>=0;j++,i--,t*=10)
{
sum+=(ch[i]-'0')*t;
}
b.a[count]=sum;
count++;
}
b.len =count++;
return in;

}
ostream& operator<<(ostream& out, BigNum& b) //重载输出运算符
{
int i;
cout << b.a[b.len - 1];
for(i = b.len - 2 ; i >= 0 ; i--)
{
cout.width(DLEN);
cout.fill('0');
cout << b.a[i];
}
return out;
}

BigNum BigNum::operator+(const BigNum & T) const //两个大数之间的相加运算
{
BigNum t(*this);
int i,big; //位数
big = T.len > len ? T.len : len;
for(i = 0 ; i < big ; i++)
{
t.a[i] +=T.a[i];
if(t.a[i] > MAXN)
{
t.a[i + 1]++;
t.a[i] -=MAXN+1;
}
}
if(t.a[big] != 0)
t.len = big + 1;
else
t.len = big;
return t;
}
BigNum BigNum::operator-(const BigNum & T) const //两个大数之间的相减运算
{
int i,j,big;
bool flag;
BigNum t1,t2;
if(*this>T)
{
t1=*this;
t2=T;
flag=0;
}
else
{
t1=T;
t2=*this;
flag=1;
}
big=t1.len;
for(i = 0 ; i < big ; i++)
{
if(t1.a[i] < t2.a[i])
{
j = i + 1;
while(t1.a[j] == 0)
j++;
t1.a[j--]--;
while(j > i)
t1.a[j--] += MAXN;
t1.a[i] += MAXN + 1 - t2.a[i];
}
else
t1.a[i] -= t2.a[i];
}
t1.len = big;
while(t1.a[t1.len - 1] == 0 && t1.len > 1)
{
t1.len--;
big--;
}
if(flag)
t1.a[big-1]=0-t1.a[big-1];
return t1;
}

BigNum BigNum::operator*(const BigNum & T) const //两个大数之间的相乘运算
{
BigNum ret;
int i,j,up;
int temp,temp1;
for(i = 0 ; i < len ; i++)
{
up = 0;
for(j = 0 ; j < T.len ; j++)
{
temp = a[i] * T.a[j] + ret.a[i + j] + up;
if(temp > MAXN)
{
temp1 = temp - temp / (MAXN + 1) * (MAXN + 1);
up = temp / (MAXN + 1);
ret.a[i + j] = temp1;
}
else
{
up = 0;
ret.a[i + j] = temp;
}
}
if(up != 0)
ret.a[i + j] = up;
}
ret.len = i + j;
while(ret.a[ret.len - 1] == 0 && ret.len > 1)
ret.len--;
return ret;
}
BigNum BigNum::operator/(const int & b) const //大数对一个整数进行相除运算
{
BigNum ret;
int i,down = 0;
for(i = len - 1 ; i >= 0 ; i--)
{
ret.a[i] = (a[i] + down * (MAXN + 1)) / b;
down = a[i] + down * (MAXN + 1) - ret.a[i] * b;
}
ret.len = len;
while(ret.a[ret.len - 1] == 0 && ret.len > 1)
ret.len--;
return ret;
}
int BigNum::operator %(const int & b) const //大数对一个int类型的变量进行取模运算
{
int i,d=0;
for (i = len-1; i>=0; i--)
{
d = ((d * (MAXN+1))% b + a[i])% b;
}
return d;
}
BigNum BigNum::operator^(const int & n) const //大数的n次方运算
{
BigNum t,ret(1);
int i;
if(n<0)
exit(-1);
if(n==0)
return 1;
if(n==1)
return *this;
int m=n;
while(m>1)
{
t=*this;
for( i=1;i<<1<=m;i<<=1)
{
t=t*t;
}
m-=i;
ret=ret*t;
if(m==1)
ret=ret*(*this);
}
return ret;
}
bool BigNum::operator>(const BigNum & T) const //大数和另一个大数的大小比较
{
int ln;
if(len > T.len)
return true;
else if(len == T.len)
{
ln = len - 1;
while(a[ln] == T.a[ln] && ln >= 0)
ln--;
if(ln >= 0 && a[ln] > T.a[ln])
return true;
else
return false;
}
else
return false;
}
bool BigNum::operator >(const int & t) const //大数和一个int类型的变量的大小比较
{
BigNum b(t);
return *this>b;
}

void BigNum::print() //输出大数
{
int i;
cout << a[len - 1];
for(i = len - 2 ; i >= 0 ; i--)
{
cout.width(DLEN);
cout.fill('0');
cout << a[i];
}
cout << endl;
}

int n, k, a[105];
BigNum dp[105][105];
int main(int argc, char const *argv[])
{
while(scanf("%d%d", &n, &k) != EOF) {
for(int i = 1; i <= n; ++i)
scanf("%d", &a[i]);
for(int i = 1; i <= k; ++i)
for(int j = 1; j <= n; ++j)
dp[i][j] = BigNum(0);
for(int i = 1; i <= n; ++i)
dp[1][i] = BigNum(1);
for(int i = 2; i <= k; ++i)
for(int j = 2; j <= n; ++j)
for(int k = 1; k < j; ++k)
if(a[j] > a[k]) dp[i][j] = dp[i][j] + dp[i - 1][k];
BigNum ans = BigNum(0);
for(int i = 1; i <= n; ++i)
ans = ans + dp[k][i];
ans.print();
}
return 0;
}