URAL1353——DP——Milliard Vasya's Function

Description

Vasya is the beginning mathematician. He decided to make an important contribution to the science and to become famous all over the world. But how can he do that if the most interesting facts such as Pythagor’s theorem are already proved? Correct! He is to think out something his own, original. So he thought out the Theory of Vasya’s Functions. Vasya’s Functions (VF) are rather simple: the value of the N^th VF in the point S is an amount of integers from 1 to N that have the sum of digits S. You seem to be great programmers, so Vasya gave you a task to find the milliard VF value (i.e. the VF with N = 10 ⁹) because Vasya himself won’t cope with the task. Can you solve the problem?

Input

Integer S (1 ≤ S ≤ 81).

Output

The milliard VF value in the point S.

Sample Input

input	output
1	10

 大意：问你从1到10^9之内所有位数相加为S的数有多少

我们定义dp[i][j]表示一共i位的数的和为j，那么可以得到状态转移方程dp[i][j] = dp[i-1][j] + dp[i-1][j-1] + ...dp[i-1][j-9]当然满足j-9 >= 0 因为考虑的是当前状态与原来状态的改变，把状态看做是一个整体，要插入的只是一个元素，dp[i-1][j]表示往入0，其他的意思就是插入k，考虑当s = 1时，要加1

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[10][85];
int main()
{
    int S;
    memset(dp,0,sizeof(dp));
    for(int i = 1; i <= 9; i++)
        dp[1][i] = 1;
    for(int i = 2; i <= 9 ;i++){
        for(int j = 1; j <= 81; j++) {
            dp[i][j] = dp[i-1][j];
                for(int k = 1; k <= 9 &&  k <= j ;k++){
                    dp[i][j] += dp[i-1][j-k];
                }
        }
    }
    while(~scanf("%d",&S)){
        int ans = 0;
        if(S == 1) {printf("10\n"); continue;}
        for(int i = 1; i <= 9;i++)
            ans += dp[i][S];
        printf("%d\n",ans);
    }
    return 0;
}