hdu 3709 Balanced Number(数位dp)

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#数据结构与算法

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Problem Description

A balanced number is a non-negative integer that can be balanced if a pivot is placed at some digit. More specifically, imagine each digit as a box with weight indicated by the digit. When a pivot is placed at some digit of the number, the distance from a digit to the pivot is the offset between it and the pivot. Then the torques of left part and right part can be calculated. It is balanced if they are the same. A balanced number must be balanced with the pivot at some of its digits. For example, 4139 is a balanced number with pivot fixed at 3. The torqueses are 4*2 + 1*1 = 9 and 9*1 = 9, for left part and right part, respectively. It's your job
to calculate the number of balanced numbers in a given range [x, y].

Input

The input contains multiple test cases. The first line is the total number of cases T (0 < T ≤ 30). For each case, there are two integers separated by a space in a line, x and y. (0 ≤ x ≤ y ≤ 10¹⁸).

Output

For each case, print the number of balanced numbers in the range [x, y] in a line.

Sample Input

0 9

7604 24324

Sample Output

897

题意:求l~r之前的平衡数

思路:枚举平衡点进行数位dp 注意每次枚举都会多算0 所以要减去多算的个数

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define ll long long int
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[2][2]={1,0 ,0,1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
int bits[20];
ll dp[20][20][2000]; //位数 平衡点位置  左右两边的和的差 
ll dfs(int len,int pos,int sum,bool ismax){
    if(!len) return sum==0; //如果左右相加为0就是平衡数 
    if(sum<0) return 0; //剪枝 
    if(!ismax&&dp[len][pos][sum]>=0) return dp[len][pos][sum];
    int up=ismax?bits[len]:9;
    ll cnt=0;
    for(int i=0;i<=up;i++){
        cnt+=dfs(len-1,pos,sum+(len-pos)*i,ismax&&i==up);
    }
    if(!ismax) dp[len][pos][sum]=cnt;
    return cnt;
}
ll solve(ll x){
    int len=0;
    while(x){
        bits[++len]=x%10;
        x/=10;
    }
    ll ans=0;
    for(int i=1;i<=len;i++){ //枚举分割点 
        ans+=dfs(len,i,0,true);
    }
    return ans-len-1;
}
int main(){
    ios::sync_with_stdio(false);
    int t;
    cin>>t;
    memset(dp,-1,sizeof(dp));
    while(t--){
        ll x,y; cin>>x>>y;
        cout<<solve(y)-solve(x-1)<<endl;
    }
    return 0;
}