标签：style blog io ar os 使用 sp for strong
Problem E
Distinct Subsequences
Input: standard input<style=‘mso-bidi-font-weight:normal‘>
Output: standard output
A subsequence of a given sequence is just the given sequence with some elements (possibly none) left out. Formally, given a sequenceX=x1x2…xm, another sequence Z=z1z2…zk is a subsequence of X if there exists a strictly increasing sequence <i1, i2, …, ik> of indices of X such that for all j = 1, 2, …, k, we have xij=zj. For example, Z=bcdb is a subsequence of X=abcbdab with corresponding index sequence < 2, 3, 5, 7 >.
In this problem your job is to write a program that counts the number of occurrences of Z in X as a subsequence such that each has a distinct index sequence.
Input
The first line of the input contains an integer N indicating the number of test cases to follow.
The first line of each test case contains a string X, composed entirely of lowercase alphabetic characters and having length no greater than 10,000. The second line contains another string Z having length no greater than 100 and also composed of only lowercase alphabetic characters. Be assured that neither Z nor any prefix or suffix of Z will have more than 10100 distinct occurrences in X as a subsequence.
Output
For each test case in the input output the number of distinct occurrences of Z in X as a subsequence. Output for each input set must be on a separate line.
Sample Input
2
Sample Output
5
____________________________________________________________________________________
Rezaul Alam Chowdhury
解题报告：
dp[i][j]表示：b的前i个字符在a的前ij个字符中出现的次数。
如果a[i] == b[j]则 dp[i][j] = dp[ii][ji-1] + dp[i-1][j-1]
如果a[i] != b[j]则 dp[i][j] = dp[i][j-1]
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
const int MAXN = 110; //数组越大，时间复杂度越高啊，谨慎使用。
char X[10010], Y[110];
struct bign
{
int len, s[MAXN];
bign()
{
memset(s, 0, sizeof(s));
len = 1;
}
bign operator = (const char *num)
{
len = strlen(num);
for(int i = 0; i < len; i++) s[i] = num[len-i-1] - ‘0‘;
return *this;
}
bign operator = (const int num)
{
char s[MAXN];
sprintf(s, "%d", num);
*this = s;
return *this;
}
bign (int num) { *this = num;}
bign (char *num) { *this = num;}
bign operator + (const bign &b)
{
bign c;
c.len = 0;
for(int i = 0, g = 0; g || i < max(len, b.len); i++)
{
int x = g;
if(i < len) x += s[i];
if(i < b.len) x += b.s[i];
c.s[c.len++] = x % 10;
g = x / 10;
}
return c;
}
bign operator - (const bign &b)
{
bign c;
c.len = 0;
for(int i = 0; i < max(len, b.len); i++)
{
c.s[i] += s[i]-b.s[i];
if(c.s[i] < 0)
{
c.s[i] += 10;
c.s[i+1]--;
}
}
while(c.s[len-1] == 0 && len > 1) len--;
c.len = len;
return c;
}
bign operator += (const bign &b)
{
*this = *this + b;
return *this;
}
string str() const
{
string res = "";
for(int i = 0; i < len; i++) res = char(s[i] + ‘0‘) + res;
if(res == "") res = "0";
return res;
}
}d[110][10010];
istream& operator >> (istream& in, bign &x)
{
string s;
in >> s;
x = s.c_str();
return in;
}
ostream& operator << (ostream& out, bign &x)
{
out << x.str();
return out;
}
void solve()
{
scanf("%s%s", X+1, Y+1);
int lenX = strlen(X+1), lenY = strlen(Y+1);
for(int i = 0; i <= lenX; i++) d[0][i] = 1;
for(int i = 1; i <= lenY; i++)
{
for(int j = i; j <= lenX; j++)
{
d[i][j] = d[i][j-1];
if(Y[i] == X[j]) d[i][j] += d[i-1][j-1];
}
}
cout<<d[lenY][lenX]<<endl;
}
int main()
{
int T;
scanf("%d", &T);
while(T--)
{
solve();
}
return 0;
}
标签：style blog io ar os 使用 sp for strong
原文地址：http://www.cnblogs.com/a972290869/p/4106989.html
