Hdu3079Balanced Number数位dp

枚举支点，然后就搞，记录之前的点的力矩和。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <climits>
#include <string>
#include <iostream>
#include <map>
#include <cstdlib>
#include <list>
#include <set>
#include <queue>
#include <stack>
#include<math.h>
using namespace std;
typedef long long LL;

LL up[100];LL dp[30][30][2000];
LL dfs(LL x,LL pos,LL pre,LL flag)
{
    if(x<=0) return pre==0;
    if(pre<0) return 0;
    if(!flag&&~dp[x][pos][pre]) return dp[x][pos][pre];
    LL limit= flag? up[x]: 9,ret= 0;
    for(LL i = 0;i<= limit;i++){
        ret+=dfs(x-1,pos,pre+(x - pos )* i ,flag&&i==limit) ;
    }
    return flag? ret : dp[x][pos][pre]= ret;
}
LL solve(LL x)
{
    LL len=0;
    while(x){
        up[++len]=x%10;
        x/=10;
    }
    LL ans=0;
    for(LL i = 1;i<= len;i++)
        ans+=dfs(len,i,0,1);
    return ans-len ;

}
int main()
{
    LL Icase;LL n,m;
    scanf("%d",&Icase);
    memset(dp,-1,sizeof(dp));
    while(Icase--){
        cin>>n>>m;
        cout<<solve(m) - solve(n-1) <<endl;
    }
    return 0;
}