hdu3709——Balanced Number

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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

20 97604 24324

Sample Output

/**hdu 3709   数位dp（自身平衡的数字）题目大意：求给定区间内满足自身平衡的数的个数，所谓平衡，          比如：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.解题思路：枚举支点。dp[i][j][k] i表示处理到的数位，j是支点，k是力矩和。但是要要把全是0的数排除*/
#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <cmath>#include <algorithm>#include <vector>#include <map>#include <string>#include <stack>using namespace std;typedef long long ll;#define PI 3.1415926535897932#define E 2.718281828459045#define INF 0x3f3f3f3f#define mod 100000007const int M=1005;int n,m;int cnt;int sx,sy,sz;int mp[1000][1000];int pa[M*10],rankk[M];int head[M*6],vis[M*100];int dis[M*100];ll prime[M*1000];bool isprime[M*1000];int lowcost[M],closet[M];char st1[5050],st2[5050];int len[M*6];typedef pair<int ,int> ac;//vector<int> g[M*10];ll dp[50][50][2000];int has[10500];int month[13]= {0,31,59,90,120,151,181,212,243,273,304,334,0};int dir[8][2]= {{0,1},{0,-1},{-1,0},{1,0},{1,1},{1,-1},{-1,1},{-1,-1}};void getpri(){    ll i;    int j;    cnt=0;    memset(isprime,false,sizeof(isprime));    for(i=2; i<1000000LL; i++)    {        if(!isprime[i])prime[cnt++]=i;        for(j=0; j<cnt&&prime[j]*i<1000000LL; j++)        {            isprime[i*prime[j]]=1;            if(i%prime[j]==0)break;        }    }}struct node{    int v,w;    node(int vv,int ww)    {        v=vv;        w=ww;    }};vector<int> g[M*100];string str[1000];int bit[50];ll dfs(int cur,int s,int presum,int e,int z){//presum为左右两边力矩和（一边正一边负），若为0，则表示平衡    if(cur<0) return presum==0;    if(presum<0)return 0;//力矩和为负，则后面的必然小于0  剪枝    if(!e&&!z&&dp[cur][s][presum]!=-1) return dp[cur][s][presum];    int endx=e?bit[cur]:9;    ll ans=0;    for(int i=0;i<=endx;i++){        if(z&&!i) ans+=dfs(cur-1,s,presum,e&&i==endx,1);//s是支点，整个dfs过程中不变        else ans+=dfs(cur-1,s,presum+(cur-s)*i,e&&i==endx,0);    }    if(!e&&!z) dp[cur][s][presum]=ans;    return ans;}ll solve(ll n){    int len=0;    while(n){        bit[len++]=n%10;        n/=10;    }    ll ans=0;    for(int i=0;i<len;i++){        ans+=dfs(len-1,i,0,1,1);//以i为支点的len位平衡数个数    }    return ans-(len-1);//len位的数，有可能是全0，因此只需保留一个0，剩下的如00，000，000，。。。都不满足条件}int main(){    int i,j,k,t;    ll l,r;    memset(dp,-1,sizeof(dp));    scanf("%d",&t);    while(t--)    {        //memset(dight,0,sizeof(dight));        scanf("%I64d%I64d",&l,&r);        printf("%I64d\n",solve(r)-solve(l-1));    }    return 0;}

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