hdu3709 Balanced Number

Balanced Number

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 1353 Accepted Submission(s): 591

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

10897
dp[i][j][k]表示有i位时，第j个数字做为中心,最后算得的权重为k，再用一下记忆化搜索就可以了，对于任意一个平衡数，一定只有一个中心使得其为平衡数，所以就可以枚举中心搜索，但是这样，会产生0,00,000,000...等这样的数，要最后减去就可以了！
#include <iostream>#include <stdio.h>#include <string.h>using namespace std;#define M 23__int64 pri[M],dp[M][M][2050];__int64 dfs(int pos,int cen,__int64 pre,int flag){    if(pos==0)return pre==0;    if(pre<0)return 0;    if(!flag&&dp[pos][cen][pre]!=-1)return dp[pos][cen][pre];    int u=flag?pri[pos]:9;__int64 ans=0;    for(int i=0;i<=u;i++)    ans+=dfs(pos-1,cen,pre+(__int64)(i*(pos-cen)),flag&&i==u);    return flag?ans:dp[pos][cen][pre]=ans;}__int64 solve(__int64 x){    int cnt=0;    while(x){        pri[++cnt]=x%10;x/=10;    }    __int64 ans=0;    for(int i=cnt;i>0;i--)    ans+=dfs(cnt,i,0,1);    return ans-cnt+1;}int main(){    int tcase;__int64 n,m;    scanf("%d",&tcase);    while(tcase--){        scanf("%I64d%I64d",&n,&m);        memset(dp,-1,sizeof(dp));        printf("%I64d\n",solve(m)-solve(n-1));    }    return 0;}