HDU-4722 Good Numbers 数位DP

　　题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4722

　　简单的数位DP，f[i][j][k]表示第 i 位数为 j 时余数为k的个数，然后直接找就可以了。。

//STATUS:C++_AC_31MS_412KB
#include <algorithm>
#include <iostream>
//#include <ext/rope>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cstring>
#include <cassert>
#include <cstdio>
#include <string>
#include <vector>
#include <bitset>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,102400000")
//using namespace __gnu_cxx;
//define
#define pii pair<int,int>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI acos(-1.0)
//typedef
typedef __int64 LL;
typedef unsigned __int64 ULL;
//const
const int N=25;
const int INF=0x3f3f3f3f;
const int MOD=1000000007,STA=8000010;
const LL LNF=1LL<<60;
const double EPS=1e-8;
const double OO=1e15;
const int dx[4]={-1,0,1,0};
const int dy[4]={0,1,0,-1};
const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
//Daily Use ...
inline int sign(double x){return (x>EPS)-(x<-EPS);}
template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
template<class T> inline T Min(T a,T b){return a<b?a:b;}
template<class T> inline T Max(T a,T b){return a>b?a:b;}
template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
//End

LL f[N][N][N];
LL a,b;
int num[N];
int T;

int getnum(LL n)
{
    int i,len=0;
    while(n){
        num[++len]=n%10;
        n/=10;
    }
    return len;
}

LL getans(LL n)
{
    int i,j,len,sum=0;
    LL ret=0;
    len=getnum(n);
    for(i=len;i>=1;i--){
        for(j=0;j<=num[i];j++){
            if(j==num[i] && i!=1)continue;
            ret+=f[i][j][(10-sum)%10];
        }
        sum=(sum+num[i])%10;
    }
    return ret;
}

int main(){
 //   freopen("in.txt","r",stdin);
    int i,j,k,p,q,ca=1;
    LL ansb,ansa;
    mem(f,0);f[0][0][0]=1;
    for(i=1;i<=19;i++){
        for(j=0;j<=9;j++){
            for(k=0;k<=9;k++){
                for(p=0;p<=9;p++){
                    f[i][j][k]+=f[i-1][p][(k-j+10)%10];
                }
            }
        }
    }
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d%I64d",&a,&b);

        ansb=(b>0)?getans(b):1;
        ansa=(a-1>0)?getans(a-1):a;

        printf("Case #%d: %I64d
",ca++,ansb-ansa);
    }
    return 0;
}