uva 10069 - Distinct Subsequences(大数相加+DP)

1、http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1010

2、数据好大，longlong也不对，用大数相加

dp[i][j]表示b的前i个字符在a的前j个字符中出现的次数

状态转移方程;

if(b[i-1]==a[j-1])//如果b的第i个字符和a的第j个字符相同，那么dp[i][j]=dp[i][j-1]+dp[i-1][j-1]

else

dp[i][j]=dp[i][j-1];

题目大意;

给定两个序列a,b,求出a中有多少个b序列，只要顺序相同即可， 不必要连续

题目；

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…x_m,another sequence Z = z₁z₂…z_k is a subsequence of X if there exists a strictly increasing sequence <i₁, i₂,…, i_k> of indices of Xsuch that for all j = 1,2, k,we have x_ij = z_j. 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

____________________________________________________________________________________
Rezaul Alam Chowdhury

4、AC代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char a[10005];
char b[105];
char dp[105][10005][105];
void add(char *c,char *a,char *b)
{
    int la=strlen(a);
    int lb=strlen(b);
    int lc=max(la,lb);
    memset(c,(lc+2)*sizeof(c[0]));
    for(int i=la-1,j=lc;i>=0;i--,j--)
    c[j]+=a[i]-'0';
    for(int i=lb-1,j--)
    c[j]+=b[i]-'0';
    for(int i=lc;i>0;c[i]+='0',i--)
    {
        if(c[i]>9)
        {
            c[i]-=10;
            c[i-1]++;
        }
    }
    if(!c[0])
    {
        for(int i=0;i<lc;i++)
        c[i]=c[i+1];
        c[lc]=0;
    }
    else
    c[0]='1';
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%s%s",a,b);
        int la=strlen(a);
        int lb=strlen(b);
        for(int i=0;i<=la;i++)
        strcpy(dp[0][i],"1");
        for(int i=1;i<=lb;i++)
        {
            for(int j=i;j<=la;j++)
            {
                if(b[i-1]==a[j-1])
                add(dp[i][j],dp[i-1][j-1],dp[i][j-1]);
                else
                strcpy(dp[i][j],dp[i][j-1]);
            }
        }
        printf("%s\n",dp[lb][la]);
    }
    return 0;
}

超时代码：（思路对，加上大数相加即可）

#include<stdio.h>
#include<string.h>
#define N 10005
char a[N];
char b[105];
int dp[105][N];//dp[i][j]表示子串的前i个字符在母串的前j个字符中出现的次数
int main()
{
    int t;
    scanf("%d",b);
        int la=strlen(a);
        int lb=strlen(b);
        memset(dp,sizeof(dp));
        for(int i=0;i<=la;i++)
        dp[0][i]=1;
        for(int i=1;i<=lb;i++)
        {
             for(int j=i;j<=la;j++)
             {
                 if(a[j-1]==b[i-1])
                 dp[i][j]=dp[i-1][j-1]+dp[i][j-1];
                 else
                 dp[i][j]=dp[i][j-1];
             }
        }
        printf("%d\n",dp[lb][la]);
    }
    return 0;
}