ima part torque ins long long off git ont itl

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 2 0 9 7604 24324 Sample Output 10 897 題意:求l~r之前的平衡數 思路:枚舉平衡點進行數位dp 註意每次枚舉都會多算0 所以要減去多算的個數

#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]; //位數 平衡點位置  左右兩邊的和的差 
ll dfs(int len,int pos,int sum,bool ismax){
    if(!len) return sum==0; //如果左右相加為0就是平衡數 
    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++){ //枚舉分割點 
        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;
}

hdu 3709 Balanced Number(數位dp)