UVa 10069 Distinct Subsequences(大数 DP)
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\n", 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
UVa 10069 Distinct Subsequences(大数 DP)的更多相关文章
- uva 10069 Distinct Subsequences 【dp+大数】
题目:uva 10069 Distinct Subsequences 题意:给出一个子串 x 和母串 s .求子串在母串中的不同序列的个数? 分析:定义dp[i][j]:x 的前 i 个字母在 s 的 ...
- uva 10069 Distinct Subsequences(高精度 + DP求解子串个数)
题目连接:10069 - Distinct Subsequences 题目大意:给出两个字符串x (lenth < 10000), z (lenth < 100), 求在x中有多少个z. ...
- UVA 10069 Distinct Subsequences(DP)
考虑两个字符串,我们用dp[i][j]表示字串第到i个和字符串到第j个的总数,由于字串必须连续 因此dp[i][j]能够有dp[i][j-1]和dp[i-1][j-1]递推而来,而不能由dp[i-1] ...
- 115. Distinct Subsequences (String; DP)
Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...
- Distinct Subsequences (dp)
Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...
- [LeetCode] Distinct Subsequences 不同的子序列
Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...
- LeetCode(115) Distinct Subsequences
题目 Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequen ...
- [Leetcode][JAVA] Distinct Subsequences
Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...
- Distinct Subsequences Leetcode
Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence ...
随机推荐
- 记一次在广播(BroadcastReceiver)或服务(Service)里弹窗的“完美”实践
事情是这样的,目前在做一个医疗项目,需要定时在某个时间段比如午休时间和晚上让我们的App休眠,那么这个时候在休眠时间段如果用户按了电源键点亮屏幕了,我们就需要弹出一个全屏的窗口去做一个人性化的提示,“ ...
- 练几道,继续过Hard题目
http://www.cnblogs.com/charlesblc/p/6384132.html 继续过Hard模式的题目吧. # Title Editorial Acceptance Diffi ...
- iipccsxxtnsoiq
gxspvyheuetwqgnbwmwd
- cocos2d-x-3.3rc2-003 cocos中的引用计数和自己主动释放池
点击打开链接
- OS 中文斜体 Italic Font Chinese - iOS_Girl
CGAffineTransform matrix = CGAffineTransformMake(1, 0, tanf(15 * (CGFloat)M_PI / 180), 1, 0, 0); UI ...
- 【Java】【Flume】Flume-NG启动过程源代码分析(一)
从bin/flume 这个shell脚本能够看到Flume的起始于org.apache.flume.node.Application类,这是flume的main函数所在. main方法首先会先解析sh ...
- (转载)Android自定义ProgressDialog进度等待框
Android自定义ProgressDialog进度等待框 作者:无缘公子 字体:[增加 减小] 类型:转载 时间:2016-01-11我要评论 这篇文章主要介绍了Android自定义Progress ...
- Android 开发环境安装配置手册
本文指导,如何一步步搭建Android开发平台. 1 下载软件 n JDK 1.5+ 到 http://java.sun.com/javase/downloads/index.jsp 下载 n ...
- ZOJ 2883 Shopaholic【贪心】
解题思路:给出n件物品,每买三件,折扣为这三件里面最便宜的那一件即将n件物品的价值按降序排序,依次选择a[3],a[6],a[9]----a[3*k] Shopaholic Time Limit: 2 ...
- canvas处理图片
canvas绘制图片的三种方法: drawImage(image, x, y) drawImage(image, x, y, width, height) drawImage(image, sourc ...