HDU

题目链接 ： HDU-3709

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 ≤ 10 ¹⁸).

Output

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

Sample Input

20 97604 24324

Sample Output

10897

题意：计算从l到r有多少个平衡数。即从该数的某一位分开，两边的权值相等。

题解：记录的时候注意记录两边的权值差值~不然数组会炸掉~

#include<stdio.h>#include<iostream>#include<set>#include<math.h>#include<algorithm>#include<string.h>#include<string>#include<queue>#include<vector>using namespace std;#define ll long longconst int inf = 0x3f3f3f3f;ll dp[20][20][3600];int digt[20];ll dfs(int pos , int mid , int sub , bool limit) {    if(pos < 0)        return sub == 0;    if(!limit && dp[pos][mid][sub] != -1)        return dp[pos][mid][sub];    int prepos = limit ? digt[pos] : 9;    ll ans =0 ;    for(int i = 0 ; i <= prepos ; i ++) {        ans += dfs(pos - 1 , mid , sub + (pos - mid)*i , limit&&i == prepos);    }    if(!limit)        dp[pos][mid][sub] = ans;    return ans;}ll solve(ll x) {    memset(digt , 0 , sizeof(digt));    int k = 0 ;    while(x) {        digt[k++] = x % 10;        x /= 10;    }    ll ans = 0 ;    for(int i = 0 ; i < k ; i ++) {        ans += dfs(k - 1 , i , 0 , 1);    }    return ans - k;}int main(){    //freopen("in.txt" , "r", stdin);    ll l , r ;    int t ;    memset(dp , -1 , sizeof(dp));    scanf("%d" , &t);    while(t--) {        scanf("%lld %lld" , &l , &r) ;        printf("%lld\n" , solve(r) - solve(l - 1));    }    return 0;}