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;

}