• C. Two Arrays


    You are given two integers nn and mm. Calculate the number of pairs of arrays (a,b)(a,b) such that:

    • the length of both arrays is equal to mm;
    • each element of each array is an integer between 11 and nn (inclusive);
    • aibiai≤bi for any index ii from 11 to mm;
    • array aa is sorted in non-descending order;
    • array bb is sorted in non-ascending order.

    As the result can be very large, you should print it modulo 109+7109+7.

    Input

    The only line contains two integers nn and mm (1n10001≤n≤1000, 1m101≤m≤10).

    Output

    Print one integer – the number of arrays aa and bb satisfying the conditions described above modulo 109+7109+7.

    Examples
    input
    Copy
    2 2
    
    output
    Copy
    5
    
    input
    Copy
    10 1
    
    output
    Copy
    55
    
    input
    Copy
    723 9
    
    output
    Copy
    157557417
    
    Note

    In the first test there are 55 suitable arrays:

    • a=[1,1],b=[2,2]
    • a=[1,2],b=[2,2]
    • a=[2,2],b=[2,2]
    • a=[1,1],b=[2,1]
    • a=[1,1],b=[1,1]

    可以将题目要求转换为:

    1、长度为2*m
    2、ci的取值范围为[1,n]
    3、序列c非递减

    我想了好久终于想明白dp转移方程的道理:

    dp[i][j]表示满足以下条件的序列个数:
    1、长度为i
    2、第一个数为j
    3、序列非递减,

    那么,dp[i][j]=(dp[i][j + 1] + dp[i - 1][j]) % mod

    比如在样例2中,i=2,j=10的时候,如何推出目前的序列数量呢?i=1,j=9的时候,只有{9}这种情况;i=2,j=10的时候,只有{10,10}这种情况

    i=1,j=9到i=2,j=9,只需要在i=1,j=9所构成的序列中把序列中任意一个数重复一遍即{9,9}这种情况;

    而i=2,j=10到i=2,j=9,只需要在i=2,j=9所构成的序列里,在最前面加上这个数就满足了题目的要求:{9,10}。

    #include <iostream>
    #include <vector>
    #include <algorithm>
    #include <string>
    #include <set>
    #include <queue>
    #include <map>
    #include <sstream>
    #include <cstdio>
    #include <cstring>
    #include <numeric>
    #include <cmath>
    #include <iomanip>
    #include <deque>
    #include <bitset>
    //#include <unordered_set>
    //#include <unordered_map>
    //#include <bits/stdc++.h>
    //#include <xfunctional>
    #define ll              long long
    #define PII             pair<int, int>
    #define rep(i,a,b)      for(int  i=a;i<=b;i++)
    #define dec(i,a,b)      for(int  i=a;i>=b;i--)
    #define pb              push_back
    #define mk              make_pair
    using namespace std;
    int dir1[6][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 },{ 1,1 },{ -1,1 } };
    int dir2[6][2] = { { 0,1 } ,{ 0,-1 },{ 1,0 },{ -1,0 },{ 1,-1 },{ -1,-1 } };
    const long long INF = 0x7f7f7f7f7f7f7f7f;
    const int inf = 0x3f3f3f3f;
    const double pi = 3.14159265358979;
    const int mod = 1000000007;
    const int N = 1005;
    //if(x<0 || x>=r || y<0 || y>=c)
    
    inline ll read()
    {
        ll x = 0; bool f = true; char c = getchar();
        while (c < '0' || c > '9') { if (c == '-') f = false; c = getchar(); }
        while (c >= '0' && c <= '9') x = (x << 1) + (x << 3) + (c ^ 48), c = getchar();
        return f ? x : -x;
    }
    ll gcd(ll m, ll n)
    {
        return n == 0 ? m : gcd(n, m % n);
    }
    
    const ll mo = 1e9 + 7;
    ll dp[22][1100];
    int main()
    {
        int n, m;
        cin >> n >> m;
        m <<= 1;
        for (int i = 1; i <= n; i++)
            dp[1][i] = 1;
        for (int i = 2; i <= m; i++)
            for (int j = n; j>0; j--)
                dp[i][j] = (dp[i][j + 1] + dp[i - 1][j]) % mo;
        ll ans = 0;
        for (int i = 1; i <= n; i++)
            ans = (ans + dp[m][i]) % mo;
        cout << ans << endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/dealer/p/12797324.html
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