nyoj VF 269 （DP）求数字和的个数

VF

时间限制：1000 ms  |           内存限制：65535 KB

难度：2

描述

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 theN^th VF in the pointS 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 withN = 10⁹) because Vasya himself won’t cope with the task. Can you solve the problem?

输入

There are multiple test cases.
Integer S (1 ≤ S ≤ 81).

输出

The milliard VF value in the point S.

样例输入

1

样例输出

10

//题意：

给你一个数n，让你在1-1000000000之间找出各个位上的数的和等于n的个数。

//思路：

动态转移方程：dp[i][j]=dp[i][j]+dp[i-1][j-k];

dp[i][j]表示数字为i位数的各个位上的数字和为j的数的个数。

知道这个动态转移方程题就好做了。。

#include<stdio.h>#include<string.h>#include<math.h>#include<algorithm>using namespace std;int dp[10][86];void find(){int i,j,k;for(i=1;i<10;i++)dp[1][i]=1;for(i=1;i<10;i++)for(j=1;j<=9*i;j++)for(k=0;k<=9&&k<=j;k++)dp[i][j]=dp[i][j]+dp[i-1][j-k];}int main(){find();int n,i,j;int cnt;while(scanf("%d",&n)!=EOF){cnt=0;if(n==1){printf("10\n");continue;}else{for(i=1;i<10;i++)cnt+=dp[i][n];}printf("%d\n",cnt);}return 0;}