数位dp（平衡数）

hdu3709 Balanced Number

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
2
0 9
7604 24324
Sample Output
10

897

题意：找出区间[x,y](0<=x<=y<=10^18)内平衡数的个数，所谓的平衡数，就是以这个数字的某一位为支点，另外两边的数字大小乘以力矩之和相等，即为平衡数

暴力是会炸的哟！

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;__int64 len,wei[22];__int64 dp[22][22][2005];__int64 dfs(__int64 pos,__int64 o,__int64 pre,__int64 flag){//flag代表当前数字有没有上限     if(pos==0){//已经遍历完每一位的时候         if(pre==0)return 1;        return 0;    }    if(pre<0)return 0;    if(flag==0&&dp[pos][o][pre]!=-1)return dp[pos][o][pre];    __int64 ans=0;    __int64 ed=flag?wei[pos]:9;    for(__int64 i=0;i<=ed;i++){        __int64 next=pre;        next+=(pos-o)*i;//算出下一步的力矩和         ans+=dfs(pos-1,o,next,flag&&i==ed);    }    if(!flag)dp[pos][o][pre]=ans;//如果当前没有上限，保存dp值     return ans;}__int64 solve(__int64 x){    if(x==-1)return 0;    if(x==0)return 1;    __int64 t=x;    len=0;    while(t){        wei[++len]=t%10;        t/=10;    }    cout<<len<<endl;    __int64 ans=0;    for(int i=len;i>=1;i--){        ans+=dfs(len,i,0,1);//以i为支点寻答案     }    ans-=(len-1);//减去全是0的情况     return ans;}int main(){    __int64 t,l,r;    memset(dp,-1,sizeof(dp));    scanf("%I64d",&t);    while(t--){        scanf("%I64d%I64d",&l,&r);        printf("%I64d\n",solve(r)-solve(l-1));    }    return 0;}