数位DP：SPOJ KPSUM - The Sum

KPSUM - The Sum

　　One of your friends wrote numbers 1, 2, 3, ..., N on the sheet of paper. After that he placed signs + and - between every pair of adjacent digits alternately. Now he wants to find the value of the expression he has made. Help him.

　　For example, if N=12 then +1 -2 +3 -4 +5 -6 +7 -8 +9 -1+0 -1+1 -1+2 = 5

Input

　　Each line contains one integer number N (1≤ N ≤ 1015). Last line contains 0 and shouldn't be processed. Number of lines in the input does not exceed 40.

Output

　　For every line in the input write the answer on a separate line.

Example

Input:
12
0

Output:
5

　　数位DP，表示一开始就打错了，然后乱改了一天，终于改对了。

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;
 long long Quick_pow(long long a,int k){
     long long ret=;
     while(k){if(k&)ret*=a;a=a*a;k>>=;}
     return ret;
 }

 long long Query(long long sum,int lena,int lenb){
     if(lena%){
         if(lenb%)
             return sum*Quick_pow(,lenb)+*Quick_pow(,lenb-);
         else
             return *Quick_pow(,lenb-);
     }
     else{
         if(lenb%)
             return *Quick_pow(,lenb-);
         else
             return sum*Quick_pow(,lenb);
     }
 }
 int st[],cnt;
 int me[]={,,-,,-,,-,,-,,-};
 long long Calc(long long a,int b){
     if(a==)return me[b];
     int flag=-;
     long long ret=;
     for(int i=;i<=b;i++){
         long long x=a*+i;cnt=;
         while(x){st[++cnt]=x%;x/=;}
         while(cnt){
             ret+=flag*st[cnt--];
             flag=-flag;
         }
     }
     return ret;
 }
 int num[],tot;
 long long mem[];
 int main(){
     mem[]=;
     for(int i=;i<=;i++){
         for(int j=;j<=;j++)
             mem[i]+=Query(-j,,i-);
         mem[i]+=mem[i-];
     }

     long long n;
     while(~scanf("%lld",&n)&&n){
         long long ret=,x=n,sum=;tot=;
         while(x){num[++tot]=x%;x/=;}
         for(int i=tot;i>;i--){
             if(!num[i])continue;
             if(i!=tot)ret+=Query(sum,tot-i+,i-);
             else ret+=mem[i-];
             for(int j=;j<num[i];j++)
                 ret+=Query(sum+Quick_pow(-,tot-i+)*j,tot-i+,i-);
              sum+=Quick_pow(-,tot-i+)*num[i];
         }
         ret+=Calc(n/,n%);
         printf("%lld\n",ret);
     }
     return ;
 }