递推DP URAL 1353 Milliard Vasya's Function

题目传送门

/*
    题意：1~1e9的数字里，各个位数数字相加和为s的个数
    递推DP：dp[i][j] 表示i位数字，当前数字和为j的个数
            状态转移方程：dp[i][j] += dp[i-1][j-k]，为了不出现负数
                改为：dp[i][j+k] += dp[i-1][j]
*/
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
using namespace std;

const int MAXN = 1e4 + 10;
const int INF = 0x3f3f3f3f;
int dp[10][90];

int main(void)        //URAL 1353 Milliard Vasya's Function
{
    //freopen ("F.in", "r", stdin);

    int s;
    while (scanf ("%d", &s) == 1)
    {
        memset (dp, 0, sizeof (dp));

        dp[0][0] = 1;
        for (int i=1; i<=9; ++i)
        {
            for (int j=0; j<=s; ++j)
            {
                if (dp[i-1][j])
                {
                    for (int k=0; k<=9; ++k)    dp[i][j+k] += dp[i-1][j];
                }
                
            }
        }

        if (s == 1)    dp[9][s] += 1;
        printf ("%d
", dp[9][s]);
    }

    return 0;
}