HDU - 5707 Combine String（DP）

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原创最新推荐文章于 2026-05-20 09:03:31 发布 347 阅读 AI 写同款· GEO 优化 ›

ABtest描述统计功效我们在面试数据分析时，常常会问到如何确定样本量的问题，会有一个指标：统计功效（power），或者是1-β 统计功效会影响实验的可信度，甚至会影响实验所需的样本量，要了解统计功效，首先要理解两类统计错误假如我们做了一个AB实验，且两组用户的数据均值为μ1和μ2，我们会有两个初始假设： H0：两个实验组的数据没有差别（μ1=μ2），这个假设也叫“零假设”，也叫原假设 H1：两个实验组之间的数据存在差别（μ1≠μ2），这个假设也叫“非零假设”，也叫备择假设定义两类错误：实验本身没有效应，但我们觉阅读详情

Combine String

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2208 Accepted Submission(s): 623

Problem Description

Given three strings

a ,

b and

c , your mission is to check whether

c is the combine string of

a and

b .
A string

c is said to be the combine string of

a and

b if and only if

c can be broken into two subsequences, when you read them as a string, one equals to

a , and the other equals to

b .
For example, ``adebcf'' is a combine string of ``abc'' and ``def''.

Input

Input file contains several test cases (no more than 20). Process to the end of file.
Each test case contains three strings

a ,

b and

c (the length of each string is between 1 and 2000).

Output

For each test case, print ``Yes'', if

c is a combine string of

a and

b , otherwise print ``No''.

Sample Input

      
       abc
def
adebcf
abc
def
abecdf

Sample Output

      
       Yes
No

Source

"巴卡斯杯" 中国大学生程序设计竞赛 - 女生专场

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题意：有a、b、c三个串，当且仅当a、b都为c的字串时输出yes，否则输出no

#include <bits/stdc++.h>

const int N = 2e3 + 10;
int dp[N][N];
char a[N], b[N], c[N];

int main()
{
    while(~scanf("%s %s %s",a , b, c))
    {
        memset(dp,0,sizeof(dp));
        int flag=0;
        int la = strlen(a), lb = strlen(b), lc = strlen(c);
        dp[0][0] = 1;
        for(int i = 0;i <= la; i++){
            for(int j = 0; j <= lb; j++){
                if(a[i] == c[i+j]){
                    dp[i+1][j] |= dp[i][j];
                }
                if(b[j]==c[i+j]){
                    dp[i][j+1] |= dp[i][j];
                }
            }
        }
        if(dp[la][lb] && lc == la + lb) printf("Yes\n");
        else printf("No\n");
    }
}

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