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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): 21    Accepted Submission(s): 10


    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

     类似最长公共子序列的dp。

    #include <bits/stdc++.h>
    using namespace std;
    #define ll long long
    const int maxn = 1e5 + 10;
    ll a[maxn], b[maxn];
    ll sum[maxn];
    const ll mod = 1000000007;
    ll dp[1010][1010];
    int main() {
        int n, m;
        while(~scanf("%d %d", &n, &m)) {
            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++) {
                    dp[i][j] = (((dp[i - 1][j] + dp[i][j - 1]) % mod - dp[i - 1][j - 1] + mod) + (a[i] == b[j] ? dp[i - 1][j - 1] + 1 : 0)) % mod;
                }
            }
            printf("%I64d
    ", dp[n][m] % mod);
        }
    }
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  • 原文地址:https://www.cnblogs.com/lonewanderer/p/5730065.html
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