F - 水之矢 URAL - 1353

mrxs

于 2018-04-03 17:41:56 发布

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分类专栏： ACM训练+实训+大学编程练习

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本文介绍了一个有趣的数学问题——Vasya的函数理论。该理论关注的是在给定数字和的情况下，寻找从1到N中所有可能的整数组合数量。通过动态规划方法解决这一问题，并提供了一个完整的C++实现代码。

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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?
Input
Integer S (1 ≤ S ≤ 81).
Output
The milliard VF value in the point S.
Example
input output
1
10

题目大意：
给出各个位置上的数字之和，然后问你有多少种组成情况。
dp[i][j]代表前j位的数字之和为i的个数。

#include <bits/stdc++.h>
using namespace std;
int dp[123][123];
int main()
{
    int n;
    cin>>n;
    memset(dp, 0, sizeof(dp));
    for(int i=1;i<=9;i++)
        dp[i][1] = 1;
    for(int j=1;j<=9;j++)
    {
        for(int i=1;i<=n;i++)
        {
            for(int k=0;k<=9;k++)
            {
                if(i>=k)
                dp[i][j] = dp[i][j] + dp[i-k][j-1];//j位的时候，
                //各个位置上的数字之和为i的个数，等于当前的+j-1位和为i-k;
            }
        }
    }
    long long int maxn = 0;
    for(int i=1;i<=9;i++)
    {
        maxn = maxn + dp[n][i];
    }
    if(n==1)
        cout<<10<<endl;
    else
    printf("%lld\n", maxn);
    return 0;
}