NYOJ 269 VF

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

动态规划！

如果找不到状态转移方程，直接递归求值，再打表！别怕麻烦哦！

递归代码：

#include<stdio.h>#include<string.h>#define M 1000000000__int64 num[85];void fun(__int64 a,__int64 sum,__int64 m){if(sum>M)return;__int64 i;num[m]++;for(i=0;i<=9;i++){fun(i,sum*10+i,m+i);}}int main(){__int64 n,i;memset(num,0,sizeof(num));for(i=1;i<=9;i++){fun(i,i,i);}while(~scanf("%I64d",&n)){printf("%I64d\n",num[n]);}return 0;}

打表：

AC码：

#include<stdio.h>int f[82]={1,10,45,165,495,1287,3003,6435,12870,24310,43749,75501,125565,202005,315315,478731,708444,1023660,1446445,2001285,2714319,3612231,4720815,6063255,7658190,9517662,11645073,14033305,16663185,19502505,22505751,25614639,28759500,31861500,34835625,37594305,40051495,42126975,43750575,44865975,45433800,45433800,44865975,43750575,42126975,40051495,37594305,34835625,31861500,28759500,25614639,22505751,19502505,16663185,14033305,11645073,9517662,7658190,6063255,4720815,3612231,2714319,2001285,1446445,1023660,708444,478731,315315,202005,125565,75501,43749,24310,12870,6435,3003,1287,495,165,45,9,1};int main(){int n;while(~scanf("%d",&n)){printf("%d\n",f[n]);}return 0;}

动态规划！

AC码：

#include<stdio.h>int dp[10][82];void DP(){int i,j,k;for(i=1;i<10;i++)dp[1][i]=1;for(i=1;i<10;i++){// i表示有i位数字时for(j=1;j<=9*i;j++){// j表示变化范围，当有i位数时，j的范围[1,9*i]for(k=0;k<10&&k<=j;k++){dp[i][j]+=dp[i-1][j-k];}}}}int main(){int n,i,ans;DP();while(~scanf("%d",&n)){if(n==1)printf("10\n");else{ans=0;for(i=1;i<10;i++)ans+=dp[i][n];printf("%d\n",ans);}}return 0;}