题意 求母串中子串出现的次数（长度不超过1后面100个0 显然要用大数了）
令a为子串 b为母串 d[i][j]表示子串前i个字母在母串前j个字母中出现的次数 当a[i]==b[j]&&d[i-1][j-1]!=0时 d[i][j]=d[i-1][j-1]+d[i][j-1]
(a[i]==b[j]时 子串前i个字母在母串前j个字母中出现的次数 等于 子串前i-1个字母在母串前j-1个字母中出现的次数 加上 子串前i个字母在母串前j-1个字母中出现的次数
a[i]!=b[j]时 子串前i个字母在母串前j个字母中出现的次数 等于 子串前i个字母在母串前j-1个字母中出现的次数)
懒得写大数模版就用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);
else
d[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]);
} else
d[i][j] = d[i][j - 1];
}
System.out.println(d[la - 1][lb - 1]);
}
in.close();
}
}
还有没加大数模版的C++代码
#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
", d[la][lb]);
}
return 0;
}
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