UVA 10069 Distinct Subsequences(dp + 高精度)

Problem E
Distinct Subsequences
Input: standard input
Output: standard output
 

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 Xsuch 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
题意：给定两个字符串，求b作为a的字串有几种情况。

思路：dp，类似LCS问题，i作为b的第i个字符，j作为a的第j个字符。 如果a[j] != b[i]。那么dp[i][j] = dp[i][j - 1]。如果相等。则多加上dp[i - 1][j - 1]。即dp[i][j] = dp[i][j - 1] + dp[i - 1][j - 1]。注意要用高精度加法

代码：

#include <stdio.h>
#include <string.h>
const int N = 10005, M = 105;

int t, i, j, n, m;
int A[M], B[M], C[M];
char a
, b[M], c[M], dp[M]
[M];
void add(char *a, char *b) {
memset(c, 0, sizeof(c));
memset(A, 0, sizeof(A));
memset(B, 0, sizeof(B));
memset(C, 0, sizeof(C));
int i, lena = strlen(a), lenb = strlen(b), len;
for (i = 0; i < lena; i ++)
A[lena - 1 - i] = a[i] - '0';
for (i = 0; i < lenb; i ++)
B[lenb - 1 - i] = b[i] - '0';
len = lena > lenb ? lena : lenb;
for (i = 0; i < len; i ++) {
C[i] += A[i] + B[i];
C[i + 1] += C[i] / 10;
C[i] %= 10;
}
while (C[len] == 0 && len > 0)
len --;
for (i = len; i >= 0; i --) {
c[len - i] = C[i] + '0';
}
}
int main() {
scanf("%d", &t);
while (t --) {
memset(dp, 0, sizeof(dp));
scanf("%s%s", a, b);
n = strlen(a); m = strlen(b);
for (i = 0; i <= n; i ++)
strcpy(dp[0][i], "1");
for (i = 1; i <= m; i ++) {
for (j = 1; j <= n; j++) {
strcpy(dp[i][j], dp[i][j - 1]);
if (b[i - 1] == a[j - 1]) {
add(dp[i][j], dp[i - 1][j - 1]);
strcpy(dp[i][j], c);
}
}
}
printf("%s\n", dp[m]
);
}
return 0;
}