NYOJ 269 VF(基础dp)（待补充）

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 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?

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

//题意大概为：1～1000000000之间，各位数字之和等于给定s的数的个数

每行给出一个数s(1 ≤ s ≤ 81)，求出1~10^9内各位数之和与s相等的数的个数。

//dp[i][j]+=dp[i-1][j-k];(k>=0 && k

//也可以递归打表

//今天拿出来重新想想又觉得不太对劲！我认为看k!=0  因为k=0的情况是前面的i-1总数的累加，

//如果k可以==0，那么后面就不应该加上sum在求一次总合！！，现在的理解是这个程序是对的！！！

#include
#include
#include
#include
#include
#include
#include
#include
const int INF = 0x3f3f3f3f;
using namespace std;
int n;
int dp[15][82]={0};
int main()
{
	for(int i=1; i<=9; i++)
			dp[1][i]=1;
	for(int i=2; i<=9; i++)
		for(int j=1; j<=i*9; j++)
			for(int k=0; k>n)
	{
		printf("%d\n",dp[9][n]);
	}
	return 0;
}

//以后如果想其他的方法，在来修改

#include
#include
#include
using namespace std;

int dp[10][82];
int main()
{
	int n;
	for(int i=1; i<=9; i++)
	    dp[1][i]=1;
	    for(int i=1; i<=9; i++)
	        for(int j=1; j<=9*i; j++)
		    for(int k=0; k<=9 && k>n)
	{
		if(n==1)
		{
			printf("10\n");
			continue;
		}
		else
		{
			int sum=0;
			for(int i=1; i<=9; i++)
			{
				sum+=dp[i][n];
			}
			printf("%d\n",sum);
		}
	}
	return 0;
}//ac