Poj 2282 digital dp

Digit dp, dp [I] [j] indicates the total number of j numbers ranging from 1 to I, and then counts by bit.

 

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>using namespace std;int dp[10][10];int pow(int a,int b){    int ret=1;    for(int i=1;i<=b;i++)    ret*=a;    return ret;}long long solve(int a,int tmp){    int s[11],lon=0;    while(a)    {        s[++lon]=a%10;        a/=10;    }    s[0]=0;    long long ans=0;    int cnt=0;    for(int i=lon;i>=1;i--)    {        if(tmp==0&&i==lon)        {            ans+=dp[i-1][tmp]*(s[i]-1);            int ff=1*pow(10,i-1)-1;            ans+=solve(ff,0);        }        else        ans+=dp[i-1][tmp]*s[i];        ans+=cnt*pow(10,i-1)*s[i];        if(tmp)        {            if(s[i]>tmp) ans+=pow(10,i-1);            if(s[i]==tmp) cnt++;        }        else if(i!=lon)        {            if(s[i]>tmp) ans+=pow(10,i-1);            if(s[i]==tmp) cnt++;        }    }    return ans+cnt;}int main(){    memset(dp,0,sizeof(dp));    for(int i=1;i<=9;i++)    {        for(int j=0;j<=9;j++)        dp[i][j]=dp[i-1][j]*10+pow(10,i-1);    }    int a,b;    while(scanf("%d %d",&a,&b),a||b)    {        if(a>b) swap(a,b);        for(int i=0;i<=9;i++)        {            printf("%lld ",solve(b,i)-solve(a-1,i));        }        printf("\n");    }    return 0;}