HDU5568/BestCoder Round #63 (div.2) B.sequence2 dp+高精度
sequence2
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.
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)
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:我们可以通过dp[i][j]dp[i][j]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] = sum{dp[k][j-1]}(k < i, b_k < b_i)dp[i][j]=sumdp[k][j−1](k<i,bk<bi)。 本题需要高精度。
//1085422276 #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <bitset> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> #include <stack> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) const int inf=~0u>>1; inline ll read() { ll x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } //************************************************************************************** #define MAX_L 405 //最大长度,可以修改 class bign { public: int len, s[MAX_L];//数的长度,记录数组 //构造函数 bign(); bign(const char*); bign(int); bool sign;//符号 1正数 0负数 string toStr() const;//转化为字符串,主要是便于输出 friend istream& operator>>(istream &,bign &);//重载输入流 friend ostream& operator<<(ostream &,bign &);//重载输出流 //重载复制 bign operator=(const char*); bign operator=(int); bign operator=(const string); //重载各种比较 bool operator>(const bign &) const; bool operator>=(const bign &) const; bool operator<(const bign &) const; bool operator<=(const bign &) const; bool operator==(const bign &) const; bool operator!=(const bign &) const; //重载四则运算 bign operator+(const bign &) const; bign operator++(); bign operator++(int); bign operator+=(const bign&); bign operator-(const bign &) const; bign operator--(); bign operator--(int); bign operator-=(const bign&); bign operator*(const bign &)const; bign operator*(const int num)const; bign operator*=(const bign&); bign operator/(const bign&)const; bign operator/=(const bign&); //四则运算的衍生运算 bign operator%(const bign&)const;//取模(余数) bign factorial()const;//阶乘 bign Sqrt()const;//整数开根(向下取整) bign pow(const bign&)const;//次方 //一些乱乱的函数 void clean(); ~bign(); }; #define max(a,b) a>b ? a : b #define min(a,b) a<b ? a : b bign::bign() { memset(s, 0, sizeof(s)); len = 1; sign = 1; } bign::bign(const char *num) { *this = num; } bign::bign(int num) { *this = num; } string bign::toStr() const { string res; res = ""; for (int i = 0; i < len; i++) res = (char)(s[i] + '0') + res; if (res == "") res = "0"; if (!sign&&res != "0") res = "-" + res; return res; } istream &operator>>(istream &in, bign &num) { string str; in>>str; num=str; return in; } ostream &operator<<(ostream &out, bign &num) { out<<num.toStr(); return out; } bign bign::operator=(const char *num) { memset(s, 0, sizeof(s)); char a[MAX_L] = ""; if (num[0] != '-') strcpy(a, num); else for (int i = 1; i < strlen(num); i++) a[i - 1] = num[i]; sign = !(num[0] == '-'); len = strlen(a); for (int i = 0; i < strlen(a); i++) s[i] = a[len - i - 1] - 48; return *this; } bign bign::operator=(int num) { char temp[MAX_L]; sprintf(temp, "%d", num); *this = temp; return *this; } bign bign::operator=(const string num) { const char *tmp; tmp = num.c_str(); *this = tmp; return *this; } bool bign::operator<(const bign &num) const { if (sign^num.sign) return num.sign; if (len != num.len) return len < num.len; for (int i = len - 1; i >= 0; i--) if (s[i] != num.s[i]) return sign ? (s[i] < num.s[i]) : (!(s[i] < num.s[i])); return !sign; } bool bign::operator>(const bign&num)const { return num < *this; } bool bign::operator<=(const bign&num)const { return !(*this>num); } bool bign::operator>=(const bign&num)const { return !(*this<num); } bool bign::operator!=(const bign&num)const { return *this > num || *this < num; } bool bign::operator==(const bign&num)const { return !(num != *this); } bign bign::operator+(const bign &num) const { if (sign^num.sign) { bign tmp = sign ? num : *this; tmp.sign = 1; return sign ? *this - tmp : num - tmp; } bign result; result.len = 0; int temp = 0; for (int i = 0; temp || i < (max(len, num.len)); i++) { int t = s[i] + num.s[i] + temp; result.s[result.len++] = t % 10; temp = t / 10; } result.sign = sign; return result; } bign bign::operator++() { *this = *this + 1; return *this; } bign bign::operator++(int) { bign old = *this; ++(*this); return old; } bign bign::operator+=(const bign &num) { *this = *this + num; return *this; } bign bign::operator-(const bign &num) const { bign b=num,a=*this; if (!num.sign && !sign) { b.sign=1; a.sign=1; return b-a; } if (!b.sign) { b.sign=1; return a+b; } if (!a.sign) { a.sign=1; b=bign(0)-(a+b); return b; } if (a<b) { bign c=(b-a); c.sign=false; return c; } bign result; result.len = 0; for (int i = 0, g = 0; i < a.len; i++) { int x = a.s[i] - g; if (i < b.len) x -= b.s[i]; if (x >= 0) g = 0; else { g = 1; x += 10; } result.s[result.len++] = x; } result.clean(); return result; } bign bign::operator * (const bign &num)const { bign result; result.len = len + num.len; for (int i = 0; i < len; i++) for (int j = 0; j < num.len; j++) result.s[i + j] += s[i] * num.s[j]; for (int i = 0; i < result.len; i++) { result.s[i + 1] += result.s[i] / 10; result.s[i] %= 10; } result.clean(); result.sign = !(sign^num.sign); return result; } bign bign::operator*(const int num)const { bign x = num; bign z = *this; return x*z; } bign bign::operator*=(const bign&num) { *this = *this * num; return *this; } bign bign::operator /(const bign&num)const { bign ans; ans.len = len - num.len + 1; if (ans.len < 0) { ans.len = 1; return ans; } bign divisor = *this, divid = num; divisor.sign = divid.sign = 1; int k = ans.len - 1; int j = len - 1; while (k >= 0) { while (divisor.s[j] == 0) j--; if (k > j) k = j; char z[MAX_L]; memset(z, 0, sizeof(z)); for (int i = j; i >= k; i--) z[j - i] = divisor.s[i] + '0'; bign dividend = z; if (dividend < divid) { k--; continue; } int key = 0; while (divid*key <= dividend) key++; key--; ans.s[k] = key; bign temp = divid*key; for (int i = 0; i < k; i++) temp = temp * 10; divisor = divisor - temp; k--; } ans.clean(); ans.sign = !(sign^num.sign); return ans; } bign bign::operator/=(const bign&num) { *this = *this / num; return *this; } bign bign::operator%(const bign& num)const { bign a = *this, b = num; a.sign = b.sign = 1; bign result, temp = a / b*b; result = a - temp; result.sign = sign; return result; } bign bign::pow(const bign& num)const { bign result = 1; for (bign i = 0; i < num; i++) result = result*(*this); return result; } bign bign::factorial()const { bign result = 1; for (bign i = 1; i <= *this; i++) result *= i; return result; } void bign::clean() { if (len == 0) len++; while (len > 1 && s[len - 1] == '\0') len--; } bign bign::Sqrt()const { if(*this<0)return -1; if(*this<=1)return *this; bign l=0,r=*this,mid; while(r-l>1) { mid=(l+r)/2; if(mid*mid>*this) r=mid; else l=mid; } return l; } bign::~bign() { } int a[101]; bign dp[111][111]; int n,k; int main() { while(cin>>n>>k) { for(int i=1; i<=n; i++) { cin>>a[i]; } bign aa=0,bb=1; for(int i=0;i<=n;i++) { for(int j=0;j<=k;j++) dp[i][j]=aa; } for(int i=1;i<=n;i++)dp[i][1]=bb; for(int i=2;i<=n;i++) { for(int j=1;j<i;j++) { if(a[i]>a[j]) { for(int kk=2;kk<=k;kk++) { dp[i][kk]+=dp[j][kk-1]; } } } } bign ans; ans=dp[1][k]; for(int i=2;i<=n;i++)ans+=dp[i][k]; cout<<ans<<endl; } return 0; }