1353. Milliard Vasya's FunctionTime 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 digitsS. 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?
InputInteger S (1 ≤ S ≤ 81).
OutputThe milliard VF value in the point S.
Sample
Problem Author: Denis Musin
Problem Source: USU Junior Championship March'2005
在1~10^9这些数中，找出各个位数和为s的个数。 dp[i][j]表示位数为i的时候，各个位数之和为j的个数，那么它是由两部分组成，一部分是dp[i-1][j],因为可以在它的低位补0，另一部分是dp[i-1][j-k](1<=k<=9)，因为dp[i-1][j-k+k]就是dp[i][j]要求的。 感觉智商被碾压~~
//0.031 206 KB
#include
#include
using namespace std;
int dp[10][107];
int main()
{
int n,j;
memset(dp,0,sizeof(dp));
for(int i=1;i<=9;i++)dp[1][i]++;//从最高位往下找，最高位只能使1~9
for(int i=2;i<=9;i++)
for(int j=1;j<=81;j++)
{
dp[i][j]=dp[i-1][j];//第一部分
for(int k=1;k<=9;k++)//第二部分
if(j-k>0)dp[i][j]+=dp[i-1][j-k];
}
while(scanf("%d",&n)!=EOF)
{
int sum=0;
if(n==1)printf("10\n");
else
{
for(int i=1;i<=9;i++)//将各个位数符合条件的加起来
sum+=dp[i][n];
printf("%d\n",sum);
}
}
return 0;
}
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