nyoj 269——VF——————【dp】

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 Nth 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 = 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


题目大意：题中给出了N。每次询问给你一个s，问你从1--N中各个位之和为s的这样的数字有多少。


解题思路：定义状态：dp[i][j]表示前i位的和为j的这样的数有多少个。i的范围只需要1-9即可，因为N的值为1e9，即使你要找i为10的数，也只有1e9这个数，所以枚举i到10是无意义的。j的范围为1-i*9，因为i位最大的和就是i位都为9。需要枚举k，k表示在第i位需要加的数字，则状态转移方程为dp[i][j]=dp[i][j]+dp[i-1][j-k]。同时需要特殊处理边界，也是最容易出差错的地方。

#include<bits/stdc++.h>
using namespace std;
const int N=1e9;
int dp[10][100];
int main(){
    int s,i,j,k,sum;
    for(i=1;i<10;i++)   //边界处理
        dp[1][i]=1;
    for(i=1;i<10;i++){
        for(j=1;j<=i*9;j++){
            for(k=0;k<10&&j>=k;k++){    //枚举第i位值k
                dp[i][j]=dp[i][j]+dp[i-1][j-k];
            }
        }
    }
    while(scanf("%d",&s)!=EOF){
        if(s==1)
            printf("10
");
        else{
            sum=0;
            for(i=1;i<10;i++)
                sum+=dp[i][s];
            printf("%d
",sum);
        }
    }
    return 0;
}