URAL 1353 Milliard Vasya's Function dp练习

最新推荐文章于 2018-03-23 16:14:27 发布

huziling

最新推荐文章于 2018-03-23 16:14:27 发布

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本文链接：https://blog.youkuaiyun.com/huziling/article/details/48032179

本文介绍了一个数学问题——Vasya函数的求解过程。该函数定义为从1到N中所有数字之和等于S的整数的数量。文章通过动态规划的方法解决了Vasya函数在N为10^9时的特殊情况。

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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?

题目弄了半天才懂，就是求10的9次以内各个位数上的数字加起来是s的数字一共有多少个

于是乎简单的dp就好

对于第 I 位总数加起来为 J 的状态如果第 i-1为是0 那么就加上dp[i-1][j] 相同于把第i-1位上的数放大第i 位并在第I位上放0，

如果i-1位不是0 那么dp[i][j] 加上 dp[i-1][j-k] k为1 - 9的每个数字

#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
int dp[11][90];
int s;
int main()
{
    for(int i = 1; i <= 10; i++)
    for(int j = 1; j <= 81; j++)
    dp[i][j] = 0;
    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++)
            {
                if(j - k >= 0)
                {
                    dp[i][j] += dp[i-1][j-k];
                }
            }
        }
    }
    while(scanf("%d",&s)!=EOF)
    {
        int ans = 0;
        if(s == 1) ans++;
        for(int i = 1; i <= 9; i++)
        {
            ans += dp[i][s];
        }
        printf("%d\n",ans);
    }
    return 0;
}