Balanced Number

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

AC代码：

#include<cstdio>#include<cstring>using namespace std;int digit[20];__int64 dp[20][20][10001];__int64 dfs(int pos,int c,int l,int lim){    if(pos==0) return l==0;    if(l<0) return 0;    if(!lim && dp[pos][c][l]!=-1) return dp[pos][c][l];    int n=lim?digit[pos]:9;    __int64 ans=0;    for(int i=0; i<=n; i++)    {        int next=l;        next+=(pos-c)*i;        ans+=dfs(pos-1,c,next,lim&&i==n);    }    if(!lim) dp[pos][c][l]=ans;    return ans;}__int64 solve(__int64 n){    int len=0;    while(n)    {        digit[++len]=n%10;        n/=10;    }    __int64 sum=0;    for(int i=1; i<=len; i++) sum+=dfs(len,i,0,1);    return sum-len+1;}int main(){    int t;    __int64 x,y;    scanf("%d",&t);    memset(dp,-1,sizeof(dp));    while(t--)    {        scanf("%I64d%I64d",&x,&y);        printf("%I64d\n",solve(y)-solve(x-1));    }    return 0;}

另一个带注释的AC代码：

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int bit[19];__int64 dp[19][19][2005];//pos为当前位置//o为支点//l为力矩//work为是否有上限__int64 dfs(int pos,int o,int l,int work){    if(pos==-1)  return l==0;//已经全部组合完了    if(l<0)  return 0;//力矩和为负，则后面的必然小于0    if(!work && dp[pos][o][l]!=-1) return dp[pos][o][l];//没有上限，且已经被搜索过了    __int64 ans=0;    int end=work?bit[pos]:9;//有上限就设为上限，否则就设为9    for(int i=0; i<=end; i++)    {        int next=l;        next+=(pos-o)*i;//力矩        ans+=dfs(pos-1,o,next,work&&i==end);    }    if(!work) dp[pos][o][l]=ans;    return ans;}__int64 solve(__int64 n){    int len=0;    while(n)    {        bit[len++]=n%10;        n/=10;    }    __int64 ans = 0;    for(int i=0; i<len; i++)  ans+=dfs(len-1,i,0,1);    return ans-(len-1);//排除掉0,00,000....这些情况}int main(){    int T;    __int64 l,r;    scanf("%d",&T);    memset(dp,-1,sizeof(dp));    while(T--)    {        scanf("%I64d%I64d",&l,&r);        printf("%I64d\n",solve(r)-solve(l-1));    }    return 0;}

View Code

hdu3709 数位dp+bfs