UVa 10069 Distinct Subsequences（大數+DP）

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Distinct Subsequences

A subsequence of a given sequence is just the given sequence with some elements (possibly none) left out. Formally, given a sequence X = 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
babgbag
bag
rabbbit
rabbit

 

Sample Output

5
3

題意 求母串中子串出現的次數（不超過1後面100個0  顯然要用大數了）
令a為子串 b為母串 d[i][j]表示子串前i個字母在母串前j個字母中出現的次數
當a[i]==b[j]&&d[i-1][j-1]!=0時 d[i][j]=d[i-1][j-1]+d[i][j-1];
(a[i]==b[j]時 子串前i個字母在母串前j個字母中出現的次數 等於 子串前i-1個字母在母串前j-1個字母中出現的次數 加上 子串前i個字母在母串前j-1個字母中出現的次數
 a[i]!=b[j]時 子串前i個字母在母串前j個字母中出現的次數 等於 子串前i個字母在母串前j-1個字母中出現的次數)
懶得寫大數模版就用java交的  ;

import java.util.*;import java.math.*;public class Main {public static void main(String args[]) {BigInteger d[][] = new BigInteger[105][10005];Scanner in = new Scanner(System.in);int t = in.nextInt();while ((t--) != 0) {String b = in.next();String a = in.next();int la = a.length();int lb = b.length();for (int i = 0; i < la; ++i)for (int j = 0; j < lb; ++j)d[i][j] = BigInteger.ZERO;if (a.charAt(0) == b.charAt(0))d[0][0] = BigInteger.ONE;for (int j = 1; j < lb; ++j) {if (a.charAt(0) == b.charAt(j))d[0][j] = d[0][j - 1].add(BigInteger.ONE);elsed[0][j] = d[0][j - 1];}for (int i = 1; i < la; ++i)for (int j = 1; j < lb; ++j) {if (a.charAt(i) == b.charAt(j)&& d[i - 1][j - 1] != BigInteger.ZERO) {d[i][j] = d[i][j - 1].add(d[i - 1][j - 1]);} elsed[i][j] = d[i][j - 1];}System.out.println(d[la - 1][lb - 1]);}in.close();}}

這裡也給上沒加大數的C++代碼 可以對比一下；

#include<cstdio>#include<cstring>using namespace std;char b[10005], a[105];int d[105][10005], la, lb, t;void dp(){    memset(d, 0, sizeof(d));    for(int j = 1; j <= lb; ++j)    {        if(a[1] == b[j]) d[1][j] = d[1][j - 1] + 1;        else d[1][j] = d[1][j - 1];    }    for(int i = 2; i <= la; ++i)        for(int j = 1; j <= lb; ++j)        {            if(a[i] == b[j] && d[i - 1][j - 1])            {                d[i][j] = d[i][j - 1] + d[i - 1][j - 1];            }            else d[i][j] = d[i][j - 1];        }}int main(){    scanf("%s", &t);    while(t--)    {        scanf("%s%s", b + 1, a + 1);        la = strlen(a + 1);        lb = strlen(b + 1);        dp();        printf("%d\n", d[la][lb]);    }    return 0;}

UVa 10069 Distinct Subsequences（大數+DP）