简单的dp(dp专题）

题目链接：https://vjudge.net/contest/216347#problem/C

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.

InputThe input contains multiple test cases.

For each test case, the first line cantains two integers

N, M (1 \leq N, M \leq 1000)

OutputFor each test case, output the answer mod 1000000007.Sample Input

Sample Output

2
3

题目大意：给你两个数组，要你求出这两个数组有多少个公共子串

个人思路：说是简单dp，但是要是想不出来就很不简单了，不说废话了，这题要求出状态转移方程，方法是用面积来求

如图，当a[i]!=b[j]时，dp[i][j]=dp[i][j-1]+dp[i-1][j]-dp[i-1][j-1],因为多加了一个dp[i-1][j-1]的面积

当a[i]==b[j]时，a[i]等于b[j]本身构成一个相等子串，dp[i-1][j-1]加上a[i],b[i]也构成新的子串，所以在不想等的情况下加上这两个条件

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

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<set>
typedef long long ll;
using namespace std;
const ll INF=1e9+7;
//#define INF 1e9+7
ll dp[1100][1100];
int main()
{
    ll n,m;
    ll a[1100],b[1100];

    while(scanf("%I64d%I64d",&n,&m)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            scanf("%I64d",&a[i]);
        }
        for(int i=1;i<=m;i++)
            scanf("%I64d",&b[i]);
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(a[i]!=b[j])
                {
                    dp[i][j]=(dp[i][j-1]+dp[i-1][j]-dp[i-1][j-1]+INF)%INF;//注意括号里面可能是负数，不加上会wa
                }
                else
                {
                    dp[i][j]=(dp[i][j-1]+dp[i-1][j]+1)%INF;
                }
            }
        }
        printf("%I64d\n",dp[n][m]);
    }
    return 0;
}

简单的dp(dp专题）

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