HDU_5791_Two（简单dp）

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Two

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1051 Accepted Submission(s): 489

Problem Description

Alice gets two sequences A and B. A easy problem comes. How many pair of sequence A' and sequence B' are same. For example, {1,2} and {1,2} are same. {1,2,4} and {1,4,2} are not same. A' is a subsequence of A. B' is a subsequence of B. The subsequnce can be not continuous. For example, {1,1,2} has 7 subsequences {1},{1},{2},{1,1},{1,2},{1,2},{1,1,2}. The answer can be very large. Output the answer mod 1000000007.

Input

The input contains multiple test cases.

For each test case, the first line cantains two integers

N,M(1≤N,M≤1000). The next line contains N integers. The next line followed M integers. All integers are between 1 and 1000.

Output

For each test case, output the answer mod 1000000007.

Sample Input

3 21 2 32 13 21 2 31 2

Sample Output

Author

ZSTU

Source

2016 Multi-University Training Contest 5

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题意

给你两个数串

问你两个数串有多少子串一致

子串不一定是连续的

解题思路

首先这个问题要想到是个dp问题

dp[i][j]代表第一个串到I位第二个串到j位相同子串的个数

则dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[I-1][j-1]减掉多的部分

但是末尾如果匹配的话

那么只要加上都减掉末尾的匹配数量再加1就可以了

因为这些子串都可以加上末尾这个元素构成新的匹配，而末尾的匹配自身也构成1个答案

#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;typedef unsigned long long LL;const int M=1005;const int MO=1e9+7;LL dp[M][M];//dp[i][j] 第一个串到i位置第二个串到j位置相同的前缀个数int nu1[M],nu2[M];int main(){    //freopen("1.in","r",stdin);    int n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        for(int i=1;i<=n;i++)            scanf("%d",&nu1[i]);        for(int i=1;i<=m;i++)            scanf("%d",&nu2[i]);        memset(dp,0,sizeof(dp));        //for(int i=1;i<M;i++)        //    dp[i][0]=dp[0][i]=1;//初始化全部置零就可以了        for(int i=1;i<=n;i++)        {            for(int j=1;j<=m;j++)            {                dp[i][j]=(dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+MO)%MO;                if(nu1[i]==nu2[j])                    dp[i][j]=(dp[i][j]+dp[i-1][j-1]+1)%MO;            }        }        //for(int i=1;i<=n;i++)        //{        //    for(int j=1;j<=m;j++)        //        cout<<dp[i][j]<<" ";        //    cout<<endl;        //}        printf("%I64d\n",dp[n][m]%MO);    }    return 0;}

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