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? 輸入
There are multiple test cases. Integer S (1 ≤ S ≤ 81).
輸出
The milliard VF value in the point S.
樣例輸入
1
樣例輸出
10
動(dòng)態(tài)規(guī)劃！
如果找不到狀態(tài)轉(zhuǎn)移方程，直接遞歸求值，再打表！別怕麻煩哦！
遞歸代碼：
#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碼：
#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;
} 動(dòng)態(tài)規(guī)劃！
AC碼：
#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位數(shù)字時(shí)for(j=1;j<=9*i;j++){// j表示變化范圍，當(dāng)有i位數(shù)時(shí)，j的范圍[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;
}
