51Nod 1043 幸运号码

#include <stdio.h>
#include <algorithm>
using namespace std;

typedef long long ll;
const int mod = 1e9 + 7;
int dp[1010][10000];
// dp[i][j] : i 个数，组成总和为j 的数量

int main()
{
    int n;
    scanf("%d", &n);
    dp[0][0] = 1;
    for (int i = 0; i <= 9; i++)
        dp[1][i] = 1;
    for (int i = 2; i <= n; i++)
    {
        for (int j = 0; j <= 9 * i; j++)
        {
            int sum = 0;
            for (int k = 0; k <= 9; k++)
            {
                if (j >= k)
                    sum = (sum + dp[i - 1][j - k]) % mod;
                else
                    break;
            }
            dp[i][j] = sum;
        }
    }
    ll ans = 0;
    for (int i = 0; i <= 9 * n; i++)
        ans = (ans + (ll)dp[n][i] * (dp[n][i] - dp[n - 1][i])) % mod;
    printf("%lld
", ans);
    return 0;
}