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  • HDU 5791 Two

    Two

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
    Total Submission(s): 163    Accepted Submission(s): 72


    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(1N,M1000). 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 2
    1 2 3
    2 1
    3 2
    1 2 3
    1 2
     
    Sample Output
    2
    3
     
    Author
    ZSTU
     状态方程 dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+(a[i]==b[j]?dp[i-1][j-1]+1:0)
    可以这么理解:
    比如序列
    a :1 2
    b :1 1 2
    当i=2  j=3
    首先不考虑a[2]和b[3]是什么  那么此时dp[i][j]=dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]把  i=2  j=3带入即可
    再考虑 a[2]==b[3]时  这时 多增加了 1+dp[i-1][j-1]对,即a[2]和b[3]匹配和 a[2]b[3]和前面dp[i-1][j-1]匹配的对数。
    a[2]!=b[3]时就不用考虑增加的....
     
    注意 子序列可以不连续...一开始卡在这里了...zz
    /* ***********************************************
    Author        :guanjun
    Created Time  :2016/8/2 14:20:58
    File Name     :p511.cpp
    ************************************************ */
    #include <iostream>
    #include <cstring>
    #include <cstdlib>
    #include <stdio.h>
    #include <algorithm>
    #include <vector>
    #include <queue>
    #include <set>
    #include <map>
    #include <string>
    #include <math.h>
    #include <stdlib.h>
    #include <iomanip>
    #include <list>
    #include <deque>
    #include <stack>
    #define ull unsigned long long
    #define ll long long
    #define mod 1000000007
    #define INF 0x3f3f3f3f
    #define maxn 10010
    #define cle(a) memset(a,0,sizeof(a))
    const ull inf = 1LL << 61;
    const double eps=1e-5;
    using namespace std;
    priority_queue<int,vector<int>,greater<int> >pq;
    struct Node{
        int x,y;
    };
    struct cmp{
        bool operator()(Node a,Node b){
            if(a.x==b.x) return a.y> b.y;
            return a.x>b.x;
        }
    };
    
    bool cmp(int a,int b){
        return a>b;
    }
    ll dp[1100][1100];
    int n,m;
    int a[1100];
    int b[1100];
    int main()
    {
        #ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
        #endif
        //freopen("out.txt","w",stdout);
        while(cin>>n>>m){
            for(int i=1;i<=n;i++)scanf("%d",&a[i]);
            for(int j=1;j<=m;j++)scanf("%d",&b[j]);
            cle(dp);
            ll x;
            for(int i=1;i<=n;i++){
                for(int j=1;j<=m;j++){
                    if(a[i]==b[j])x=dp[i-1][j-1]+1;
                    else x=0;
                    dp[i][j]=(dp[i][j]+dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1]+mod)%mod;
                    dp[i][j]=(dp[i][j]+x)%mod;
                }
            }
            printf("%I64d
    ",dp[n][m]);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/pk28/p/5730375.html
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