数位DP：SPOJ KPSUM

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=1;
    while(k){if(k&1)ret*=a;a=a*a;k>>=1;}
    return ret;
}

long long Query(long long sum,int lena,int lenb){
    if(lena%2){
        if(lenb%2)
            return sum*Quick_pow(10,lenb)+45*Quick_pow(10,lenb-1);
        else
            return 5*Quick_pow(10,lenb-1);    
    }
    else{
        if(lenb%2)
            return 5*Quick_pow(10,lenb-1);
        else
            return sum*Quick_pow(10,lenb);    
    }
}
int st[20],cnt;
int me[15]={0,1,-1,2,-2,3,-3,4,-4,5,-5};
long long Calc(long long a,int b){
    if(a==0)return me[b]; 
    int flag=-1;
    long long ret=0;
    for(int i=0;i<=b;i++){
        long long x=a*10+i;cnt=0;
        while(x){st[++cnt]=x%10;x/=10;}
        while(cnt){
            ret+=flag*st[cnt--];
            flag=-flag;
        }
    }
    return ret;
}
int num[25],tot;
long long mem[25]; 
int main(){
    mem[1]=5;
    for(int i=2;i<=15;i++){
        for(int j=1;j<=9;j++)
            mem[i]+=Query(-j,1,i-1);
        mem[i]+=mem[i-1];
    }
    
    long long n;
    while(~scanf("%lld",&n)&&n){
        long long ret=0,x=n,sum=0;tot=0;
        while(x){num[++tot]=x%10;x/=10;}
        for(int i=tot;i>1;i--){
            if(!num[i])continue;
            if(i!=tot)ret+=Query(sum,tot-i+1,i-1);
            else ret+=mem[i-1];
            for(int j=1;j<num[i];j++)
                ret+=Query(sum+Quick_pow(-1,tot-i+1)*j,tot-i+1,i-1);
             sum+=Quick_pow(-1,tot-i+1)*num[i];
        }
        ret+=Calc(n/10,n%10);
        printf("%lld
",ret);
    }
    return 0;
}