Milliard Vasya's Function-Ural1353动态规划

Time limit: 1.0 second	Memory limit: 64 MB

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 Nth 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 = 109) 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	output
1	10

Problem Author: Denis Musin
Problem Source: USU Junior Championship March’2005

计算[0,10^9]之间有多少各数的各位之和为s,Dp[i][j]表示前i位中和为j的个数，则对于第i为k是它的个数为Dp[i][j]+=Dp[i-1][j-k].
因此递推式就出来了。

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <algorithm>

using namespace std;

int Dp[10][83];

int main()
{
    memset(Dp,0,sizeof(Dp));

    for(int i = 0;i<=9;i++)//初始化第一位
    {
        Dp[1][i]=1;
    }
    for(int i=2;i<=9;i++)
    {
        for(int j=0;j<=81;j++)
        {
            Dp[i][j]=  Dp[i-1][j];
        }
        for(int j=0;j<=81;j++)
        {
            for(int s = 1;s<=9&&s<=j;s++)
            {
                Dp[i][j] +=Dp[i-1][j-s];
            }
        }
    }
    int n;
    Dp[9][1]++;//由于1000000000计算不到，因此要加上
    while(~scanf("%d",&n))
    {
        printf("%d\n",Dp[9][n]);
    }
    return 0;
}